what is the expected absorbance of a standard solution made by dissolving 0.0070 mol of nicl2 * 6h20 in water to make 100 ml of solution?

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Answer 1

The expected absorbance of a standard solution made by dissolving 0.0070 mol of NiCl2 · 6H2O in water to make 100 ml of solution is 0.227.

Absorbance is a measure of the quantity of light that passes through a sample relative to the quantity of light that passes through a blank sample.

The sample absorbance is determined by the sample's concentration, thickness, and absorbing properties of the solution.

In order to calculate the expected absorbance of a standard solution made by dissolving 0.0070 mol of NiCl2 · 6H2O in water to make 100 ml of solution, we need to use the Beer-Lambert Law.

It states that the absorbance of a solution is directly proportional to the concentration of the solution and the length of the path that the light has to travel through the solution.

So, A = εlc where A = absorbanceε = molar extinction coefficient l = path length c = concentration Since the path length and molar extinction coefficient are constant, the absorbance is proportional to the concentration.

So, A1/A2 = C1/C2

Where, A1 = absorbance of the standard solutionC1 = concentration of the standard solution

A2 = absorbance of the unknown solutionC2 = concentration of the unknown solution Rearranging the formula we get, C2 = C1(A2/A1)

Given that the concentration of the standard solution is 0.0070 mol/L and the path length is 1 cm.

The molar extinction coefficient for NiCl2·6H2O is 4.76 × 10^3 L/mol·cm. Substituting these values in the formula we get, C2 = 0.0070 mol/L × (0.380/1.660) = 0.0016 mol/L

Again, using the Beer-Lambert law we can find the expected absorbance of the unknown solution, where A = εlc.A = 4.76 × 10^3 L/mol·cm × 1 cm × 0.0016 mol/L = 7.62.

The expected absorbance of a standard solution made by dissolving 0.0070 mol of NiCl2 · 6H2O in water to make 100 ml of solution is 0.227.

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Related Questions

A lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO3)2. Which of the following statements is correct? Ksp=6.44 x 10^(-3) for CdF2. A) The presence of NaF will raise the solubility of Cd(NO3)2B) The solubility of cadmium fluoride is increased by the presence of additional fluoride ions.C) One must know Ksp for cadmium nitrate to make meaningful predictions on this system. D) Cadmium fluoride precipitates until the solution is saturated. E) The solution is unsaturated and no precipitate forms. stel et shnt nan ha added to 1.00 L of

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When a lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO3)2, the correct statement is that B) The solubility of cadmium fluoride is increased by the presence of additional fluoride ions.

How does the addition of anions affect the solubility of salts?

The solubility of salts is influenced by the presence of anions.

The solubility of salts is increased by the presence of anions in some cases. Anions reduce the solubility of salts in other cases. Cadmium nitrate (Cd(NO3)2) has a Ksp of 6.44 × 10−3, which must be compared to the ion product (IP) for Cd(NO3)2 in solution to decide whether precipitation will occur. Cd(NO3)2 is a soluble salt that ionizes according to the following equation:

Cd(NO3)2 → Cd2+ + 2 NO3−.

According to the solubility product rule, the IP for Cd(NO3)2 is determined as IP = [Cd2+][NO3−]^2. Because cadmium fluoride (CdF2) is less soluble than cadmium nitrate, it must be compared to the IP for CdF2 in solution to decide whether precipitation will occur. The ion product (IP) for CdF2 in solution can be calculated using the stoichiometry of the equilibrium between Cd2+ and F− ions: Cd2+(aq) + 2F−(aq) → CdF2(s).

Thus, IP = [Cd2+][F−]^2. As a result, the addition of fluoride ions to the Cd(NO3)2 solution in the form of NaF increases the solubility of cadmium fluoride because the concentration of F− ions is increased. As a result, option B is correct.

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Astronomers studying the planet of Acer have detected igneous rock under its surface. One astronomer makes a claim that some of the material that this igneous rock formed from used to be in sedimentary rock on the surface of Acer. If the scientist is correct, how could sedimentary rock have become igneous rock?

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If the astronomer's claim is correct and igneous rock was formed from material that was originally in sedimentary rock on the surface of Acer, then the process that likely occurred is called "igneous intrusion."

What is Igneous intrusion?

Igneous intrusion happens when molten rock, known as magma, is forced into layers of sedimentary rock, which is formed from the accumulation of sediments like sand, mud, or organic matter. As the magma intrudes into the sedimentary rock, it heats up the surrounding rocks and causes them to partially melt and recrystallize. Over time, as the magma cools and solidifies, it forms igneous rock.

The process of igneous intrusion can also cause the sedimentary rock layers to fold or deform, creating features like faults, folds, and uplifts. These changes in the sedimentary rock can be used by geologists to understand the history and geology of a particular region.

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for the next several questions, use the following information: a 2.00 g sample of ammonia (nh3 ) reactants with 4.00 g of oxygen to form nitrogen monoxide and water. all of the reactants and products are gases. do not forget about diatomic molecules.

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Since we are given the reactants and products in a chemical reaction, we can write the balanced chemical equation as:

4 NH3 + 5 O2 → 4 NO + 6 H2O

From the balanced equation, we can see that 4 moles of NH3 react with 5 moles of O2 to form 4 moles of NO and 6 moles of H2O.

To solve the following questions, we can use the stoichiometry of the balanced chemical equation.

How many moles of NH3 are in the sample?

The molar mass of NH3 is 17.03 g/mol, so the number of moles of NH3 in the sample is:

2.00 g / 17.03 g/mol = 0.1173 mol NH3

How many moles of O2 are in excess?

We can first calculate the number of moles of O2 required to react completely with NH3. From the balanced equation, we know that 4 moles of NH3 react with 5 moles of O2, so the number of moles of O2 required is:

0.1173 mol NH3 × (5 mol O2 / 4 mol NH3) = 0.1466 mol O2

The actual amount of O2 used is 4.00 g / 32.00 g/mol = 0.125 mol O2, so the number of moles of O2 in excess is:

0.125 mol O2 - 0.1466 mol O2 = -0.0216 mol O2

Since the value is negative, it means that O2 is the limiting reactant, and NH3 is in excess.

How many moles of H2O are produced?

From the balanced equation, we know that for every 4 moles of NH3 reacted, 6 moles of H2O are produced. Therefore, the number of moles of H2O produced is:

0.1173 mol NH3 × (6 mol H2O / 4 mol NH3) = 0.1760 mol H2O

What is the mass of NO produced?

The molar mass of NO is 30.01 g/mol, so the mass of NO produced is:

0.1173 mol NH3 × (4 mol NO / 4 mol NH3) × 30.01 g/mol = 3.52 g NO

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Given the solubility rules from the book, which of the following metal hydroxides should be soluble in water? LiOH CuOH AgOH. Cu(OH)2 TlOH. LiOH.

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The metal hydroxide that should be soluble in water among LiOH, CuOH, AgOH, Cu(OH)₂, and TlOH is LiOH.

1. LiOH: Lithium hydroxide (LiOH) is an alkali metal hydroxide, and alkali metal hydroxides are generally soluble in water. So, LiOH is soluble.

2. CuOH: Copper(I) hydroxide (CuOH) is a transition metal hydroxide, which are typically insoluble. Therefore, CuOH is not soluble.

3. AgOH: Silver hydroxide (AgOH) is also a transition metal hydroxide and is insoluble in water.

4. Cu(OH)₂: Copper(II) hydroxide (Cu(OH)₂) is another transition metal hydroxide and is insoluble in water.

5. TlOH: Thallium hydroxide (TlOH) is also a transition metal hydroxide, and like most transition metal hydroxides, it is insoluble in water.

In conclusion, among the given metal hydroxides, LiOH is soluble in water.

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What aldehyde is needed to prepare the carboxylic acid by an oxidation reaction?

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Answer:

The oxidation of an aldehyde can be achieved using a variety of oxidizing agents, including potassium permanganate (KMnO4), chromium trioxide (CrO3), and silver oxide (Ag2O). The specific oxidizing agent used will depend on the conditions and desired yield.

For example, if we want to prepare acetic acid, we can oxidize ethanol (an alcohol) using a strong oxidizing agent like potassium permanganate. Alternatively, we can oxidize acetaldehyde (an aldehyde) using a milder oxidizing agent like silver oxide.

Therefore, any aldehyde can be used to prepare a carboxylic acid by oxidation, but the specific oxidizing agent and reaction conditions may vary depending on the aldehyde and desired yield.

The aldehyde that is need for the preparation of the acid is CH3(CH2)8CH(Cl)CHO

How do you prepare an acid from an aldehyde?

It is not possible to directly prepare an acid from an aldehyde as an aldehyde is already an oxidized form of a primary alcohol, which can be further oxidized to form a carboxylic acid.

Aldehydes can be oxidized to carboxylic acids using strong oxidizing agents such as potassium permanganate (KMnO4) or chromic acid (H2CrO4). The reaction conditions need to be carefully controlled to avoid over-oxidation of the aldehyde to carbon dioxide.

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the defense mechanism in which self-justifying explanations replace the real, unconscious reasons for actions is group of answer choices rationalization. denial. projection. reaction formation. displacement.

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Answer: The defense mechanism in which self-justifying explanations replace the real, unconscious reasons for actions is Rationalization.

Rationalization is a type of defense mechanism where individuals create a logical explanation for their own behavior, even if the behavior is actually driven by emotions or unconscious thoughts.

This type of defense is used to protect the ego from the anxiety of a certain situation, usually one that is perceived to be too uncomfortable or overwhelming.

By rationalizing a behavior, the individual is able to tell themselves that they did the right thing, even if the choice was not made consciously or with the best intentions. Rationalization is a way to protect one’s ego by creating a logical justification for an action.

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if a sample containing 6.5 moles of nh3 is reacted with excess cuo, how many moles of each product can be made? n2

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The amount of N2 (Nitrogen) produced will be limited by the amount of NH3 (Ammonia) present. Thus, the maximum amount of N2 that can be produced is 1.625 moles (which is half of the 3.25 moles calculated above). Therefore, the answer is 1.625 moles of N2.

If a sample containing 6.5 moles of NH3 is reacted with excess CuO, 1.625 moles of N2 can be produced. There are two products that can be produced by the reaction of NH3 with excess CuO: N2 and H2O. The balanced equation for this reaction is as follows: 4NH3 + 3CuO → 2N2 + 3H2O + 3CuTo determine how many moles of each product can be made, we need to use the mole ratio between NH3 and the products. From the balanced equation, we can see that for every 4 moles of NH3, 2 moles of N2 can be produced. Therefore, for 6.5 moles of NH3, we can calculate the amount of N2 produced as follows:6.5 moles NH3 × (2 moles N2/4 moles NH3) = 3.25 moles N2However, we have to remember that the reaction is carried out with excess CuO. This means that all of the NH3 will be consumed, and there will be enough CuO (Copper oxide) to react with all of it. Therefore, the amount of N2 produced will be limited by the amount of NH3 present. Thus, the maximum amount of N2 that can be produced is 1.625 moles (which is half of the 3.25 moles calculated above). Therefore, the answer is 1.625 moles of N2.

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calculate the volume in ml of a 6 m solution of hcl stock solution required to make 250 ml of 50 mm hcl?

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The volume in ml of a 6 m solution of hcl stock solution required to make 250 ml of 50 mm hcl is: 20.8 ml.

To calculate the volume of a 6 M HCl stock solution required to make 250 ml of 50 mM HCl, use the following equation:

volume of stock solution (ml) = (desired concentration (mM) x volume of desired solution (ml)) / stock solution concentration (M).

Therefore, in this case, volume of stock solution (ml) = (50 mM x 250 ml) / 6 M = 20.8 ml. In other words, 20.8 ml of a 6 M HCl stock solution is required to make 250 ml of 50 mM HCl. This is because the number of moles (the amount of HCl molecules) in the solution must remain constant.

Increasing the volume of the solution by dilution means that the concentration (the amount of HCl molecules per ml of solution) must be decreased, and thus the amount of HCl stock solution must be increased.

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When Pt metal is used as a catalyst for the previous reaction, we see that the mechanism changes and the reaction is much faster. The activation energy is found to be 98.4 kJ mol-1 with the catalyst at room temperature. How much would you have to raise the temperature to get the catalyzed reaction to run 100 times faster than it does at room temperature with the catalyst? Please answer in °C.

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The temperature should be raised by 28.15°C to run 100 times faster than it does at room temperature with the catalyst.

How to find temperature of a catalytic reaction?

To determine the temperature increase needed to make the catalyzed reaction run 100 times faster, we can use the Arrhenius equation:

[tex]k_{2}[/tex]/[tex]k_{1}[/tex] = e^(-Ea/R * (1/[tex]T_{2}[/tex] - 1/[tex]T_{1}[/tex])

Where [tex]k_{1}[/tex] and [tex]k_{2}[/tex] are the rate constants at temperatures [tex]T_{1}[/tex] and [tex]T_{2}[/tex], Ea is the activation energy (98.4 kJ mol-1), and R is the gas constant (8.314 J [tex]K^{-1}[/tex] [tex]mol^{-1}[/tex]).

Since we want the reaction to be 100 times faster, k2/k1 = 100. Now we can rearrange the equation and solve for [tex]T_{2}[/tex]:

1/[tex]T_{2}[/tex] - 1/[tex]T_{1}[/tex] = -R * ln(100)/Ea

Assuming room temperature ([tex]T_{1}[/tex]) is 298 K (25°C), we can plug in the values:

1/[tex]T_{2}[/tex] - 1/298 = -8.314 * ln(100)/98,400

1/[tex]T_{2}[/tex] = 1/298 + (8.314 * ln(100)/98,400)

[tex]T_{2}[/tex] = 1 / (1/298 + (8.314 * ln(100)/98,400))

Now, calculate the value of [tex]T_{2}[/tex]:

[tex]T_{2}[/tex] ≈ 326.3 K

To convert [tex]T_{2}[/tex] to °C, subtract 273.15:

[tex]T_{2}[/tex] = 326.3 - 273.15 ≈ 53.15°C

Therefore, you would need to raise the temperature by approximately 28.15°C (53.15 - 25) to make the catalyzed reaction run 100 times faster.

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assuming ideal behavior, how many liters hcl gas are required to make concentrated hydrochloric acid (11.6 mol/l) at 25oc and 1 atm pressure?

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520.67 liters of HCl gas are required to make concentrated hydrochloric acid (11.6 mol/L) at 25°C and 1 atm pressure. while assuming ideal behavior.

To make concentrated hydrochloric acid (11.6 mol/L) at 25°C and 1 atm pressure, the volume of HCl gas needed is 520.67 L.

Assuming ideal behavior,

Molarity (M) = number of moles of solute/volume of solution in liters (L)

Given:

Molarity (M) = 11.6 mol/L

Volume of solution (V) = ?

Temperature (T) = 25°C

Pressure (P) = 1 atm

We can use the ideal gas law to find the volume of HCl gas required to make 1 L of concentrated HCl. Then, we can use this value to find the volume of HCl gas required to make a certain volume of concentrated HCl. The ideal gas law is given as:

PV = nRT

where: P is pressure, V is volume of the gas, n is the number of moles of gas, R is the gas constant, T is the temperature. We can rearrange the ideal gas law to solve for volume:

V = nRT/PAt

standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 L.

Therefore, the number of moles of HCl gas required to make 1 L of concentrated HCl is given as:

11.6 mol/L × 1 L = 11.6 moles

We can substitute these values into the ideal gas law equation and solve for the volume of HCl gas required to make 1 L of concentrated HCl:

V = nRT/PV = (11.6 mol) × (0.08206 L·atm/K·mol) × (298 K)/(1 atm)V

= 260.51 L

However, we are interested in finding the volume of HCl gas required to make a certain volume of concentrated HCl. We can use the following conversion factor to find the volume of HCl gas required:

1 L concentrated HCl = 260.51 L HCl gas

We can use dimensional analysis to solve for the volume of HCl gas required to make 1 L of concentrated HCl:

11.6 mol/L × 1 L concentrated HCl × (260.51 L HCl gas/1 L concentrated HCl) = 3020.37 L HCl gas

However, this calculation gives the volume of HCl gas required to make 1 L of concentrated HCl.

We are interested in finding the volume of HCl gas required to make a certain amount of concentrated HCl.

We can use the following formula to solve for the volume of HCl gas required to make a certain amount of concentrated HCl:

V2 = V1 × (M1/M2)

where:V1 is the volume of concentrated HCl needed

M1 is the molarity of concentrated HCl

M2 is the molarity of the HCl gas

V2 is the volume of HCl gas needed

We can substitute the given values into the formula and solve for

V2:V2 = (1 L) × (11.6 mol/L)/(0.08206 L·atm/K·mol × 298 K)V2

= 520.67 L

Therefore, 520.67 liters of HCl gas are required to make concentrated hydrochloric acid (11.6 mol/L) at 25°C and 1 atm pressure.

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How many chlorine atoms are there in 4 molecules of HCl?

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Answer: Hydrogen chloride is a diatomic molecule, consisting of a hydrogen atom H and a chlorine atom Cl connected by a polar covalent bond.

if you dissolve .750 moles of sodium sulfate in .500 liters of soltuion, what is the total concentration, in moles/liter, of the sodium ions present in solution

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Total concentration of sodium ions is 3.00 moles/liter.

The concentration of sodium ions in a solution containing 0.750 moles of sodium sulfate dissolved in 0.500 liters of solvent can be determined by first finding the number of moles of sodium ions present in the solution.

The sodium ions are derived from the dissociation of sodium sulfate in water, which produces two moles of sodium ions for every mole of sodium sulfate. Since there are 0.750 moles of sodium sulfate in the solution, there are 1.5 moles of sodium ions present in the solution.

To calculate the total concentration of sodium ions, divide the number of moles of sodium ions by the volume of the solution in liters:Total concentration of sodium ions = moles of sodium ions / liters of solution

Total concentration of sodium ions = 1.5 moles / 0.500 liters = 3.00 moles/liter

Therefore, the total concentration of sodium ions present in the solution is 3.00 moles/liter.

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How do you calculate the number of electrons in each energy level sugar has

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Answer: the formula 2n2 :)

Explanation: To calculate the maximum number of electrons in each energy level, the formula 2n2 can be used, where n is the principal energy level (first quantum number). For example, energy level 1, 2(1)2 calculates to two possible electrons that will fit into the first energy level.

I hope it helped! :)

what is the mass of sodium chloride required to create a 0.875 m solution 534 g of water. how many moles of nacl is required

Answers

The mass of sodium chloride that is required to create a 0.875 M solution 534 g of water is 27.291 g and 0.467 moles of NaCl is required.

Mass of water = 534 g

Molality of the solution = 0.875 m

Molality is the number of moles of solute per kilogram of solvent.

It is represented by the formula:

Molality = number of moles of solute / kilogram solvent

Its mathematical expression is:

m = n/kg

Now we will convert the g into kg.

Mass of water = 534 g× 1kg/1000 g = 0.534 kg

putting the values in formula:

0.875 m = n / 0.534 kg

n = 0.467 mol

Now we will calculate the mass of sodium chloride:

Mass = number of moles × molar mass

Mass = 0.467 mol × 58.44 g/mol

Mass = 27.291 g

Thus, the required mass and moles of NaCl are 27.291g and 0.467mol respectively.

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A face-centered cubic unit cell is the repeating unit in which type of crystal packing:__________

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A face-centered cubic unit cell is the repeating unit in which type of crystal packing: cubic closest-packed, option B.

Solids can be thought of as having a structure similar to that of a piece of wallpaper in three dimensions. Wallpaper has a recurring pattern that is consistent and runs from edge to edge. Similar repeating patterns may be found in crystals, however in this case, the patterns span three dimensions from one edge of the solid to the other.

By describing the dimensions, form, and content of the most basic repeating unit in the pattern, we may accurately describe a piece of wallpaper. The smallest repeating unit's dimensions, composition, and arrangement on top of one another to form the crystal may be used to characterise a three-dimensional crystal.

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Complete question:

A face-centered cubic unit cell is the repeating unit in which type of crystal packing A) hexagonal close-packing B)cubic close-packed C)body centered D)simple E)all of the above

a sample is sent to the laboratory for an anti-xa assay. the result of the ptt is 65.7 seconds. the result of the anti-xa assay is 0.9 u/ml of heparin. the patient is on lovenox. their anti-xa level is:

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b. Therapeutic. For treatment dosage therapy, the therapeutic anti-Xa level is between 0.5 and 1 units/mL. For prophylactic dosage treatment, the ideal anti-Xa level is between 0.2 and 0.4 units/ml.

The activity of heparin, including low molecular weight heparin, is measured using the anti-Xa assay. Anti Xa is an ambiguous name. Heparin activity is what the lab truly reports when it says "against Xa." Therefore, low anti-Xa correlates with lower heparin activity, whereas high Xa correlates with higher heparin activity. The medicine and the indication both affect the therapeutic anti-Xa activity. Unfractionated heparin has a different range than low molecular weight heparin. For the treatment of venous thromboembolism, a therapeutic range for unfractionated heparin is 0.35–0.7 and for low molecular weight heparin, it is 0.5–1. 10% less is the suggested goal for acute coronary syndrome.

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Complete Question:

A sample is sent to the laboratory for an anti-Xa assay. The result of the PTT is 65.7 seconds. The result of the anti-Xa assay is 0.9 U/mL of heparin. The patient is on Lovebox. Their anti-Xa level is:

a. subtherapeutic

b. therapeutic

c. supratherapeutic

d. prophylactic

a mixture of gases can be described as a solution because it is a homogeneous mixture that has a uniform composition throughout at the molecular level TRUE/FALSE

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The given statement "a mixture of gases can be described as a solution because it is a homogeneous mixture that has a uniform composition throughout at the molecular level" is true because  properties of the mixture are the same throughout, and the composition of the mixture does not vary from one part to another.

A mixture of gases can be described as a solution because it is a homogeneous mixture, meaning that the composition is uniform throughout the mixture. This is true at the molecular level because the gases are thoroughly mixed, and the molecules of each gas are distributed evenly throughout the mixture.

Therefore, the properties of the mixture are the same throughout, and the composition of the mixture does not vary from one part to another.

Thus the given statement  "a mixture of gases can be described as a solution because it is a homogeneous mixture that has a uniform composition throughout at the molecular level" is true.

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presented with two tubes: one tube with a buffered solution + acid and one tube with water + acid, how will you know which tube has the buffer and which tube does not have the buffer?

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One tube has a buffered solution + acid and the other tube has water + acid. To decide whether or not the solution is buffered, a simple pH test can be done. An acid-base indicator can be used to determine the pH of each solution.

A buffered solution is defined as a solution that can withstand minor changes in pH upon the addition of small amounts of an acid or base.

Consider the following steps:

To both tubes, add a small amount of acid-base indicator. Determine the pH of each solution by observing the color change of the acid-base indicator when it is added to it. The pH of the solution is determined by the color of the acid-base indicator after it has been added to it. Compare the pH of the two solutions. The solution with the lower pH is likely to have a buffer, whereas the solution with the higher pH is unlikely to have a buffer. This is due to the fact that the addition of an acid to a buffered solution would result in a lower pH, whereas the addition of an acid to an unbuffered solution would result in a higher pH. To find out which tube has the buffer and which does not, one has to compare the pH of each solution.

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a compound containing only carbon and hydrogen has a carbon to hydrogen ratio of 11.89 . which carbon to hydrogen mass ratio is possible for another compound composed only of carbon and hydrogen?

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Answer : Another compound composed of only carbon and hydrogen can have any carbon to hydrogen mass ratio, depending on the number of atoms in the molecule and the atomic weights of the elements.

A compound containing only carbon and hydrogen can have any carbon to hydrogen mass ratio. This is because each element has its own atomic weight, and when combined in a compound the ratio of atoms or molecules can be different from the ratios of elements. For example, methane (CH4) has a mass ratio of 12:1 (carbon to hydrogen), while ethane (C2H6) has a mass ratio of 6:3.

It is important to note that the mass ratio is not the same as the molar ratio, which is determined by the number of atoms in the molecule. For example, ethylene (C2H4) has a molar ratio of 1:2, but its mass ratio is 6:4.

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A 0.682-gram sample of an unknown weak monoprotic organic acid, HA, was dissolved in sufficient water to make 50.0 mL of solution and was titrated with a 0.135 M NaOH solution. After the addition of 10.6 mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 27.4 mL of the 0.135 M NaOH.
a. Calculate the number of moles of acid in the original sample.
b. Calculate the molar mass of the organic acid.
c. Calculate the molarity of the unreacted HA remaining in the solution at pH = 5.65.

Answers

a. The number of moles of acid in the original sample is 0.00369. b. The molar mass of the organic acid is 0.135  M. c. The molarity of the unreacted HA remaining in the solution at pH 5.65 is 0.045 M

Calculation:

a. The equivalence point was reached after the addition of 27.4 mL of the 0.135 M NaOH.a.

Moles of NaOH = M × V = 0.135 M × 27.4 mL = 0.00369 moles

Using the balanced equation, we find that the number of moles of HA is equal to the number of moles of NaOH at the equivalence point. HA + NaOH → NaA + HOH0. 00369 moles of NaOH are needed to react with 0.00369 moles of HA.

b. Molar mass of HA = (mass of HA) / (number of moles of HA) = 0.682 g / 0.00369 moles = 184.7 g/molc. Calculate the molarity of the unreacted HA remaining in the solution at pH = 5.65.The pH of the solution was 5.65 after 10.6 mL of NaOH were added.

c. To calculate the molarity of the remaining HA, we first need to find the pKa of the acid.

pH = pKa + log([A-]/[HA])5.65 = pKa + log([A-]/[HA]). We know that at the equivalence point, [A-] = [HA] / 2.

Therefore,[A-] = 0.00369 moles / 2 = 0.00185 moles[Ligand] = (moles of ligand) / (liters of solution). We need to find [HA] in moles/L, so we need to find [A-] in moles/L. We can use the molarity of the NaOH solution to do this. [NaOH] = 0.135 M

moles of NaOH = [NaOH] × (liters of solution)moles of NaOH = 0.135 M × 0.0106 L.

moles of NaOH = 0.00144 moles

moles of HA at pH = 5.65 = moles of HA initially - moles of NaOH added = 0.00369 moles - 0.00144 moles

= 0.00225 moles[HA] = 0.00225 moles / 0.050 L = 0.045 M

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What would you predict, the solubility of KHT (solid) in pure water compared with the solubility of KHT (solid) in a 0.1 M KCl solution, which one will be higher? Explain your answer.

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The solubility of KHT (solid) in pure water compared with the solubility of KHT (solid) in a 0.1 M KCl solution is predicted to be higher in the 0.1 M KCl solution. This is because the KCl solution has a higher ionic strength, increasing the solubility of ionic compounds like KHT.

Let's understand this in detail:

What is solubility?

Solubility is defined as the ability of a substance to dissolve in a particular solvent under certain conditions. It measures the maximum amount of solute that can be dissolved in a given amount of solvent at a particular temperature, pressure, and other conditions.

Solubility of KHT in pure water:

KHT (Potassium hydrogen tartrate) is a weak acid salt that has low solubility in pure water. The solubility of KHT in pure water is affected by various factors such as temperature, pH, and pressure. The solubility of KHT in pure water is around 4.4 g/L at room temperature.

Solubility of KHT in 0.1 M KCl solution: The solubility of KHT in a 0.1 M KCl solution is predicted to be higher than in pure water. KCl is an ionic salt dissociating in water to produce K+ and Cl- ions. The presence of KCl increases the ionic strength of the solution. This ionic strength improves the solubility of other ionic compounds, such as KHT. KHT has a higher solubility in a 0.1 M KCl solution than in pure water due to this reason.

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how many ml of 0.280 m barium nitrate are required as bariium sulfate all the sulfate ions from 25.0 ml of 0.350 m aluminum sulfate?

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To determine how many milliliters (ml) of 0.280 m barium nitrate are required to remove all of the sulfate ions from 25.0 ml of 0.350 m aluminum sulfate, you can use the following equation:

Molarity (M) = moles/volume (V)

First, calculate the number of moles of sulfate ions in the given volume of aluminum sulfate.

M = 0.350 M = moles/25.0 ml

moles = 0.350 M x 25.0 ml = 8.75 moles

Next, calculate the number of moles of barium nitrate that are needed to completely remove the sulfate ions.

M = 0.280 M = moles/V

moles = 8.75 moles/V

V = 8.75 moles/0.280 M = 31.25 ml

Therefore, 31.25 ml of 0.280 m barium nitrate is required to remove all of the sulfate ions from 25.0 ml of 0.350 m aluminum sulfate.

This is because molarity (M) is a measure of concentration that is equal to moles of a substance divided by the volume of the solution (V). Thus, to remove the sulfate ions from the aluminum sulfate solution, you must calculate the molarity of the aluminum sulfate, calculate the number of moles of sulfate ions in the solution, and then calculate the number of moles of barium nitrate that are needed to completely remove the sulfate ions. The volume of barium nitrate required is equal to the number of moles of sulfate ions divided by the molarity of the barium nitrate.  

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what is the percent by weight (w/w%) of sugar in soda? assume the average mass of sugar in soda is 31.0 g and the total mass is 370.0 g.

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The percent by weight (w/w%) of sugar in soda, assuming the average mass of sugar in soda is 31.0 g and the total mass is 370.0 g, is 8.38%.

The mass percent composition of a compound is a measure of the ratio of the mass of each component to the total mass of the compound. It is denoted by w/w%.

The mass percentage of a component in a solution can be calculated using the following formula:

the mass percent of a component = (mass of the component ÷ total mass of solution) × 100

Assume the average mass of sugar in soda is 31.0 g and the total mass is 370.0 g.

To determine the weight percentage of sugar in soda, the mass percent composition formula can be used as follows:

mass percent of sugar = (mass of sugar ÷ total mass of soda) × 100

mass percent of sugar = (31.0 g ÷ 370.0 g) × 100

mass percent of sugar = 0.0838 × 100

mass percent of sugar = 8.38%

Therefore, the percent by weight (w/w%) of sugar in soda, assuming the average mass of sugar in soda is 31.0 g and the total mass is 370.0 g, is 8.38%.

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What is one way that the layers of the atmosphere help to maintain life on Earth?

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One way that the layers of the atmosphere help to maintain life on Earth is by absorbing and scattering harmful solar radiation, such as ultraviolet (UV) radiation.

The ozone layer, which is located in the stratosphere layer of the atmosphere, absorbs most of the Sun's harmful UV radiation, preventing it from reaching the Earth's surface where it can cause DNA damage and skin cancer. Additionally, the atmosphere helps regulate the Earth's temperature by trapping heat from the Sun through the greenhouse effect, which is essential for maintaining a stable and habitable climate. The atmosphere also contains oxygen, which is necessary for the survival of many living organisms.

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if 37.2 kj of energy is evolved when 100. g of glucose is fermented, what is the molar enthalpy of fermentation?

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If 37.2 kJ of energy is evolved when 100g. So, the molar enthalpy of fermentation is 67 kJ/mol.

The molar enthalpy of fermentation can be calculated as follows:

From the equation, 1 mole of glucose yields 2 moles of ethanol and 2 moles of carbon dioxide. Thus, the balanced equation for this process is:

C₆H₁₂O₆ (aq)  → 2C₂H₅OH(aq) + 2CO₂ (g)

From the given values, the mass of glucose that was fermented is 100 g. The molar mass of glucose is 180.16 g/mol. Thus, the number of moles of glucose can be calculated as follows:

moles of glucose = Mass of glucose / Molar mass of glucose

moles of glucose = 100 g / 180.16 g/mol

moles of glucose = 0.555 moles

The molar enthalpy of fermentation is defined as the amount of energy released per mole of fermented glucose. Thus, the molar enthalpy of fermentation can be calculated as follows:

Molar enthalpy  = Energy released / moles of glucose

Molar enthalpy  = 37.2 kJ / 0.555 mol

Molar enthalpy  = 67 kJ/mol

Therefore, the molar enthalpy of fermentation is 67 kJ/mol.

Complete question:

The equation for the fermentation of glucose to ethanol and carbon dioxide is C6 H12 O6 (aq) 3,2CrN 5 OH(aq)+2CO 2 (g) If 37.2 kJ of energy is evolved when 100. g of glucose is fermented, what the molar enthalpy of fermentation?

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The major product of the synthesis, 3-methoxyheptane, is produced by an SN2 mechanism. Complete the reaction mechanism below by adding curved arrows showing electron flow and the final product, including the correct stereochemistry, to demonstrate the formation of the major product (3-methoxyheptane).

Answers

To demonstrate the formation of 3-methoxyheptane through an SN2 mechanism, follow these steps:

1. Identify the nucleophile and electrophile: The nucleophile is the methoxide ion (CH3O-) and the electrophile is the alkyl halide, such as 1-chloroheptane (C7H15Cl).

2. Show the electron flow using curved arrows: Draw a curved arrow from the lone pair on the oxygen atom of the methoxide ion to the carbon atom bonded to the chlorine in 1-chloroheptane. This arrow represents the nucleophilic attack.

3. Show the leaving group departure: Draw another curved arrow from the carbon-chlorine bond in 1-chloroheptane to the chlorine atom. This arrow represents the departure of the chloride ion (Cl-) as the leaving group.

4. Draw the final product with the correct stereochemistry: As SN2 reactions lead to inversion of stereochemistry, if the starting 1-chloroheptane had an R configuration, the final product, 3-methoxyheptane, would have an S configuration (and vice versa). So, draw the final product with the methoxy group (OCH3) attached to the third carbon atom of the heptane chain, and the correct stereochemistry based on the starting material.

The resulting structure will be 3-methoxyheptane, with the appropriate stereochemistry.

a certain combustion reaction generates 4.50 moles of carbon dioxide how many grams does this represent report your answer to 3 significant figures

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If any combustion reaction generates 4.50 moles of carbon dioxide then the equivalant amount in grams will be 198 g (in 3 significant figures).

The molar mass of carbon dioxide (CO2) is qual to 44.01 g/mol.

In order to find the mass of 4.50 moles of CO2, we can use the following formula,

mass = number of moles × molar mass

Substituting the provided values, we will obtain,

mass = 4.50 mol × 44.01 g/mol

mass = 198.045 g

Therefore, after rounding to three significant figures, the mass of 4.50 moles of CO2 is obtaine to be 198 g.

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If any combustion reaction generates 4.50 moles of carbon dioxide then the equivalant amount in grams will be 198 g (in 3 significant figures).

The molar mass of carbon dioxide (CO2) is qual to 44.01 g/mol.

In order to find the mass of 4.50 moles of CO2, we can use the following formula,

mass = number of moles × molar mass

Substituting the provided values, we will obtain,

mass = 4.50 mol × 44.01 g/mol

mass = 198.045 g

Therefore, after rounding to three significant figures, the mass of 4.50 moles of CO2 is obtaine to be 198 g.

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Order the anionic compounds from the most basic to least basic? 3) 1) 2) Magnify Select One Strongest base Second strongest Match With B 3) Weakest base C 2)

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The compounds containing anions from the most basic to least basic are:1) B (Strongest base)2) C3) A (Weakest base)The order of basicity of anionic compounds can be determined using the periodic table. The correct answer is B>C>A.

Anions are larger than their corresponding atoms due to the addition of one or more electrons. As a result, anions have lower effective nuclear charges and therefore are more basic than their parent atoms. The larger the anion, the more basic it is. The order of basicity of anionic compounds is as follows:

B > C > A

Where, B is the most basic anionic compound, C is the second most basic anionic compound, A is the least basic anionic compound

Therefore, the order of the anionic compounds from the most basic to least basic is B > C > A. To order the anionic compounds from the most basic to least basic, follow these steps: Identify the anions present in each compound., Determine the conjugate acid of each anion, Compare the strength of the conjugate acids, Order the anionic compounds based on the strength of their conjugate acids (the weaker the conjugate acid, the stronger the base).

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the student then prepares a solution using four pellets of naoh dissolved to 100.00 ml in a volumetric flask. the student slowly adds this to the khp solution to perform a titration. it requires 22.50 ml of the naoh solution to reach the endpoint. what is the molarity of the naoh solution based on this titration?

Answers

The volume of NaOH solution used in the titration is 22.50 mL or 0.0225 L. The molarity of the NaOH solution is 0.210 mol/L.

To determine the molarity of the NaOH solution, we can use the balanced chemical equation for the reaction between NaOH and KHP:

NaOH + KHP → NaKP + H2O

From the equation, we can see that one mole of NaOH reacts with one mole of KHP. Therefore, the number of moles of NaOH used in the titration can be calculated by:

moles NaOH = molarity of NaOH solution × volume of NaOH solution used (in liters)

The volume of NaOH solution used in the titration is 22.50 mL or 0.0225 L.

To calculate the molarity of the NaOH solution, we need to determine the number of moles of NaOH used in the titration. From the balanced equation, we can see that one mole of KHP reacts with one mole of NaOH. The mass of KHP used in the titration is 0.969 g, which corresponds to the number of moles of KHP used:

moles KHP = mass of KHP / molar mass of KHP

= 0.969 g / 204.22 g/mol

= 0.004738 mol

Since the stoichiometry of the reaction is 1:1, the number of moles of NaOH used in the titration is also 0.004738 mol. Substituting these values into the above equation, we get:

0.004738 mol = molarity of NaOH solution × 0.0225 L

Solving for the molarity of the NaOH solution, we get:

molarity of NaOH solution = 0.004738 mol / 0.0225 L

= 0.210 mol/L

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if a 9.48g of hydrosulfuric acid is reacte with 6.35g of silver nitrate calculate the mass in g of solid silver sulifde formed

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Therefore, the mass of silver sulfide formed when 9.48 g of hydrosulfuric acid is reacted with 6.35 g of silver nitrate is 2.238 g.

When 9.48 g of hydrosulfuric acid is reacted with 6.35 g of silver nitrate, the reaction forms solid silver sulfide. The equation for this reaction is:

H₂S + 2 AgNO₃ → Ag₂S + 2 HNO₃.

To calculate the mass of silver sulfide formed, we need to use the mole ratio of the two reactants. We know that the molecular weight of silver nitrate is 169.88 g/mol and the molecular weight of hydrosulfuric acid is 34.08 g/mol.

Using the mole ratio, we can find the moles of each reactant:

9.48 g/34.08 g/mol = 0.2786 moles of H₂S and 6.35 g/169.88 g/mol = 0.0373 moles of AgNO₃.

Since the reaction forms 1 mole of Ag₂S for every 2 moles of AgNO3, we can calculate the moles of Ag₂S formed: (0.0373 moles of AgNO₃ x 1 mole of Ag₂S)/2 moles of AgNO₃ = 0.01865 moles of AgS.

Now, using the molecular weight of silver sulfide (119.97 g/mol), we can calculate the mass of silver sulfide formed: 0.01865 moles of Ag₂S x 119.97 g/mol = 2.238 g of Ag₂S.


Therefore, the mass of silver sulfide formed when 9.48 g of hydrosulfuric acid is reacted with 6.35 g of silver nitrate is 2.238 g.

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