Answer:
the pressure at the top of the mountain was 0.85 atm.
Explanation:
We can use the combined gas law to solve this problem:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.
Assuming that the temperature remains constant, we can simplify the combined gas law to:
P1 * V1 = P2 * V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
Substituting the given values, we get:
P1 * 1.0 L = 1 atm * 0.85 L
P1 = (1 atm * 0.85 L) / 1.0 L
P1 = 0.85 atm
How many moles do 235.5 liters of carbon dioxide
Gas legislation
PV = nRT
where P is atmospheric pressure
Volume (measured in litres)
The number of moles is n.
R = gas constant (0.0821 L atm/(mol K))
Temperature (in Kelvin) equals T.
We need to rearrange the ideal gas law to solve for n:
n = PV / RT
To use this equation, we need to know the pressure, temperature, and volume of the carbon dioxide.
0.002355 moles of carbon dioxide are equal to 235.5 litres of the petrol. As a result, 0.002355 moles of carbon dioxide are equivalent to 0.053 litres.
This is because 1 mole of carbon dioxide is equal to 22.4 litres. This indicates that there are 0.002355 moles of carbon dioxide for every 0.053 litres of carbon dioxide.
Therefore, there are 0.002355 moles of carbon dioxide in 235.5 litres of carbon dioxide.
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According to the law of conservation of mass, how much zinc was produced if Calcium = 25 g, Zinc Carbonate = 125 g, and Calcium Carbonate = 95 g.
HELPP
All the balanced chemical equations obey the law of conservation of mass. The numbers which are used to balance the chemical equation are called the coefficients. So here the mass of zinc is 55 g.
According to the law of conservation of mass, the mass can neither be created nor be destroyed but it can be converted from one form to another. The reactants appear on the left hand side and the products appear on the right hand side.
The amount of products is equal to the amount of reactants according to the law of conservation of mass.
25 + 125 = 95 + Zn
Zn = 55 g
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Calculate the percentage composition of elements in the following compounds:
a. Water: H₂O
b. Glucose: C6H12O6
c. Calcium nitrate: Ca(NO3)2
d. Aluminum sulfate: Al2(SO4)3
e. Magnesium phosphate: Mg3(PO4)2
The percentage composition of Hydrogen (H) in water (H₂O) is 11.1%
Breakdown of How Percentage Composition is CalculatedPercentage composition is finding the amount of individual elements that made up a compound.
To calculate, we use the formula:
% composition = [tex]\frac{Atomic Mass}{Molar Mass} x 100[/tex]
(a) Water: H₂O
To calculate the percentage composition of elements in water, we need to find the molar mass of water, which is:
Molar mass of H₂O = (2 × atomic mass of H) + (1 × atomic mass of O)
Molar mass of H₂O = (2 × 1.0 g/mol) + (1 × 16.00 g/mol)
Molar mass of H₂O = 18.0 g/mol
%Hydrogen = 2/18 x100 = 11.11
%Oxygen: 16/18 x 100 = 88.89%
b. Glucose: C₆H₁₂O₆
Molar mass of C₆H₁₂O₆ = 6 (12) + 12(1) + 6(16)
= 72 + 12 + 96
= 180g/mol
Now, we can calculate the percentage composition of elements in glucose:
% Carbon: 72/180 = 40%
% Hydrogen: 12/180 = 7%
Percentage composition of oxygen: (6 × atomic mass of O) / molar mass of C6H12O6 × 100%
Percentage composition of oxygen: (6 × 16.00 g/mol) / 180.18 g/mol × 100%
Percentage composition of oxygen: 53.29%
Therefore, the percentage composition of elements in glucose is 39.99% carbon, 6.72% hydrogen, and 53.29% oxygen.
c. Calcium nitrate: Ca(NO3)2
To calculate the percentage composition of elements in calcium nitrate, we need to find the molar mass of calcium nitrate, which is:
Molar mass of Ca(NO3)2 = atomic mass of Ca + (2 × atomic mass of N) + (6 × atomic mass of O)
Molar mass of Ca(NO3)2 = 40.08 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)
Molar mass of Ca(NO3)2 = 164.09 g/mol
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A certain flexible weather balloon contains 3.1 L of helium gas. Initially, the balloon is in WP at 8500ft, where the temperature is 23.8oC and the barometric pressure is 564.8 torr. The balloon then is taken to the top of Pike’s Peak at an altitude of 14,100ft, where the pressure is 400 torr and the temperature is 6.9oC. What is the new volume of the balloon at the top of Pikes Peak?
The concept combined gas law is used here to determine the new volume of the balloon. This law relate one thermodynamic variable to another holding everything else constant. The new volume is 4.12 L.
The combination of Boyles law, Charles's law and Avogadro's law gives the combined gas law. This law states that the product of pressure volume and temperature of a system remains constant.
The equation is:
P₁V₁ / T₁ = P₂V₂ / T₂
V₂ = P₁V₁T₂ / P₂T₁
0.7455 × 3.1 × 279.9 / 0.528 × 296.8 = 4.12 L
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Reaction Ai Sodium Bicarbonate and Hydrochloric Acid
Experimental Data
(a) Mass of evaporating dish
watch glass
(b) Mass of evaporating dish watch glass sodium bicarbonate
(c) Mass of sodium bicarbonate used
(d) Mass of evaporating dish watch glass sodium chloride
(e) Mass of sodium chloride collected (experimental yield)
.
Mole Ration and Reaction Story
.
.
Data Analysis
1) Use your data to determine the experimental mole-to-mole ratio between sodium bicarbonate and sodium chloride
Show your work for each
NaHCOS
Convert the mass of sodium bicarbonate used to moles
100.69
1
mole
9
Convert the mass of sodium chloride collected in moder
Nac
2g
104.2
3.bg
mole
g
Divide both of your results from the preceding two steps by the lower mole value to determine the simplest mole-to-
mole ratio between sodium bicarbonate and sodium chloride.
The reaction between sodium carbonate and hydrochloric acid
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
How to solveFor reaction A
Mass of sodium bicarbonate used = (Mas of evaporating dish + watch glas + sodium bicarbonate) - (Mas of evaporating dish + watch glas)
= 46.582 - 46.263
= 0.319 g
Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)
= 46.473 - 46.263
= 0.210 g
Moles of sodium bicarbonate (NaHCO3) used = mas/molecular weight
= (0.319 g) / (84 g/mol)
= 0.00380 mol
Moles of sodium chloride (NaCl) used = mas/molecular weight
= (0.210 g) / (58.44 g/mol)
= 0.00359 mol
Mol ratio of NaHCO3 : NaCl = 0.00380 : 0.00359
Divide by 0.00359
Simple mol ratio
NaHCO3 : NaCl = 1.06 : 1
After rounding
Mol ratio of NaHCO3 : NaCl = 1 : 1
Moles of NaHCO3 = moles of NaCl = 0.00359 mol
Theoretical yield of NaCl = moles x molecular weight
= 0.00359 mol x 58.44 g/mol
= 0.210 g
the percent yield of sodium chloride
= actual yield x 100 / theoretical yield
= 0.210*100/0.210
= 100%
the reaction between sodium bicarbonate and hydrochloric acid
NaHCO3 + HCl = NaCl + CO2 + H2O
For reaction B
Mass of sodium carbonate used = (Mas of evaporating dish + watch glas + sodium carbonate) - (Mas of evaporating dish + watch glas)
= 51.677 - 51.368
= 0.309 g
Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)
= 51.671 - 51.368
= 0.303 g
Moles of sodium carbonate (Na2CO3) used = mas/molecular weight
= (0.309 g) / (106 g/mol)
= 0.00292 mol
Moles of sodium chloride (NaCl) used = mas/molecular weight
= (0.303 g) / (58.44 g/mol)
= 0.00518 mol
Mol ratio of
Na2CO3 : NaCl = 0.00292 : 0.00518
Divide by 0.00292
Simple mol ratio
Na2CO3 : NaCl = 1 : 1.78
After rounding
Mol ratio of Na2CO3 : NaCl = 1 : 2
Moles of NaCl = 2 x moles of Na2CO3
= 2 x 0.00292 = 0.00584 mol
Theoretical yield of NaCl = moles x molecular weight
= 0.00584 mol x 58.44 g/mol
= 0.341 g
the percent yield of sodium chloride
= actual yield x 100 / theoretical yield
= 0.303*100/0.341
= 88.86%
the reaction between sodium carbonate and hydrochloric acid
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
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a 70 piece of metal at 120 C is dropped into a kilometer with 150 g of 30C water the final temperature of the water and little changes 35c what is the specific heat of the metal
Answer:
cm=0.385 J
cm = (0.15 kg x 4.18 J/gCx (35C - 30C)) / (70 pieces x 0.1 kg/piece x (120C - 35C))
The answer is:
cm = 0.385 J/gc
Therefore, the specific heat of the metal is 0.385
An OBJECT absorbs like between the light wavelengths of 430 - 400 nm. What is the color of the OBJECT?
The color of the object appears as yellow/orange.
What is the color?
The visible light spectrum spans a wavelength range of 400 to 700 nanometers (nm), with shorter wavelengths appearing as blue or violet and longer wavelengths as red.
This object is absorbing light with wavelengths between 430 and 400 nm, indicating that it is absorbing light that is visible in the blue and violet spectrum.
We know that this color that we see is actually a complementary color to blue/violet from the color wheel.
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Using six carbon atoms as an example, write the condensed structural formula and the names of the following functional groups: alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide.
For alcohol, ether, aldehyde, ketone, carboxylic acid, ester, amine and amide, the condensed structural formulas of six carbon members of the series are shown below.
What is the condensed structural formula?The condensed structural formula for an alcohol with six carbon atoms is C6H13OH
The condensed structural formula for an ether with six carbon atoms is C6H14O
The condensed structural formula for an aldehyde with six carbon atoms is C6H10O
The condensed structural formula for a ketone with six carbon atoms is C6H10O
The condensed structural formula for a carboxylic acid with six carbon atoms is C6H10O2
The condensed structural formula for an ester with six carbon atoms is C6H12O2
The condensed structural formula for an amine with six carbon atoms is C6H15N
The condensed structural formula for an amide with six carbon atoms is C6H13NO
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PLEASE HELP NOW
Caffeine is a weak base with a b of 4.1×10^-4 Calculate the initial molar concentration of a solution of caffeine if the pH is 10.94.
Answer:430 mg/L = 0.43g/L
Explanation:
what is the name for CH3-CH2-C(O)-OCH3
Answer:
The name for CH3-CH2-C(O)-OCH3 is ethyl methanoate.
Explanation:
Make a drawing representing the reaction that occurs between calcium nitrate and sodium oxalate.
The reaction between calcium nitrate (Ca(NO₃)₂) and sodium oxalate (Na₂C₂O₄) can be represented by the following chemical equation:
Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2NaNO₃
This is a double displacement reaction, where the calcium ion (Ca²⁺) from calcium nitrate and the oxalate ion (C₂O₄²⁻) from sodium oxalate switch places to form calcium oxalate (CaC₂O₄) and sodium nitrate (NaNO₃). The balanced chemical equation shows that one mole of calcium nitrate reacts with one mole of sodium oxalate to form one mole of calcium oxalate and two moles of sodium nitrate.
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Potential in a different kind of cell.
A typical mammalian cell at 37
∘
C, with only potassium channels open, will have the following equilibrium:
K+ (intracellular) ⇌ K+ (extracellular),
with an intracellular concentration of 150 mM K+, and 4.0 mM K+ in the extracellular fluid.
What is the potential, in volts, across this cell membrane? Note: in this case, n = the charge on the ion, and Eo for a concentration cell = 0.00 V. explain please
The potential across this cell membrane with only potassium channels open is -0.082 V, which means that the inside of the cell is negatively charged relative to the outside.
The potential across a cell membrane can be calculated using the Nernst equation:
E = (RT/zF) ln([ion]out/[ion]in)
E = potential in volts, R= gas constant (8.314 J/mol*K), T= temperature in Kelvin, z = charge on the ion, F= Faraday constant (96,485 C/mol), and [ion]out and [ion]in are the concentrations of the ion outside and inside the cell, respectively.
K+ (intracellular) ⇌ K+ (extracellular)
The charge on potassium ions is +1, so z = 1.
The temperature is 37°C or 310 K.
The concentrations of potassium ions are [K+]in = 150 mM and [K+]out = 4.0 mM.
Substituting these values into the Nernst equation,
E = (RT/zF) ln([K+]out/[K+]in)
E = (8.314 J/mol.K × 310 K)/(1 × 96,485 C/mol) ln(4.0 mM/150 mM)
E = -0.082 V
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What is the molarity 10.0g of Cr(NO3)3 in 325 mL of solution
Answer:
Explanation:
molar mass Cr(NO3)3 = 238 g/mol
Convert 325 ml to liters: 325 mls x 1 L / 1000 mls = 0.325 L
Convert 10.0 g to moles: 10.0 g x 1 mol / 238 g = 0.0420 moles
Molarity = moles/liters = 0.0420 moles / 0.325 L = 0.129 M (3 sig. figs.)
How should the Key change for a weak base?
We've figured out what part of the salt causes the flame to change color, so now let's measure the wavelengths created with four metals.
Use the ruler under the "tools" icon in the upper right of the video player to measure the wavelengths of light released by each compound.
The wavelength of one of the spectral lines for strontium chloride SrCl₂ is approximately 460.7 nanometers (nm).
When strontium chloride SrCl₂ is heated, it emits a characteristic red color, which indicates that it produces spectral lines in the red part of the visible spectrum. The most intense spectral line for SrCl₂ is at approximately 460.7 nm, which corresponds to the transition from the 5² electronic configuration to the 4d state.
This transition is responsible for the red color observed when strontium chloride is introduced to a flame. The wavelength of a spectral line is related to the energy of the transition and is given by
λ = hc ÷ E
where λ is the wavelength, h is Planck's constant, c is the speed of light, and E is the energy of the transition. In the case of SrCl₂, the energy of the transition from 5s² to 4d is approximately 2.69 eV, which corresponds to a wavelength of 460.7 nm.
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The complete question is:
What is the wavelength of one of the spectral lines for strontium chloride SrCl₂?
You have 900,000 atoms of a radioactive substance. After 4 half-lives have past, how many atoms remain?
you cannot have a fraction of an atom, so round the answer to the nearest whole number.
The number of atoms remaining after 4 half-lives can be calculated using the formula: N = N0 /[tex]2^4[/tex] . Therefore, after 4 half-lives, approximately 56,250 atoms of the radioactive substance remain.
Radioactive decay is the process by which a nucleus of an atom loses energy by emitting ionizing radiation. The rate of decay of a radioactive substance is measured by its half-life, which is the time it takes for half of the radioactive atoms in a sample to decay.
N = N0 /[tex]2^n[/tex]
where: N0 = initial number of atoms N = final number of atoms
Substituting the given values,
N = 900,000 / [tex]2^4[/tex]
N = 56,250
Rounding to the nearest whole number,
N ≈ 56,250 atoms
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(25 pts) Explain in depth the relationship between LIMITING reactant and THEORETICAL yield
Answer: I got you fam
Explanation:
A limiting reactant is a reactant in a chemical reaction that limits the amount of product created.
So for example if there are elements X and Y reacting to create product XY, once say element X runs out, the reaction stops, even though there is still more of the reactant Y. So there is 0 g of element X remaining, and maybe 2 g left of element Y. X is the limiting reactant since it limits the amount of product made.
Theoretical yield is the maximum amount of product that could be made in an experiment. This occurs if a reaction is 100% effective (and experimentally, this doesn't usually happen, which is why it is called theoretical).
In Activity 2, you tested various compounds for chemical changes. (Barium nitrate, sodium hydroxide, sodium hydrogen carbonate, copper (II) sulfate, potassium iodide, silver nitrate, iron (III) nitrate, and hydrochloric acid.) Write the chemical formulas for each of the reactants.
The chemical formulas are as follows:
Barium nitrate: Ba(NO₃)₂Sodium hydroxide: NaOHSodium hydrogen carbonate: NaHCO₃Copper (II) sulfate: CuSO₄Potassium iodide: KISilver nitrate: AgNO₃Iron (III) nitrate: Fe(NO₃)₃Hydrochloric acid: HClChemical formulas are shorthand notations used to represent the composition of a substance. In this case, the reactants used in Activity 2 are listed with their chemical formulas.
Barium nitrate is represented by the chemical formula Ba(NO₃)₂, which shows that it contains one barium ion (Ba²⁺) and two nitrate ions (NO₃⁻).
Sodium hydroxide is represented by the chemical formula NaOH, which shows that it contains one sodium ion (Na⁺) and one hydroxide ion (OH⁻).
Sodium hydrogen carbonate is represented by the chemical formula NaHCO₃, which shows that it contains one sodium ion (Na⁺), one hydrogen ion (H⁺), one carbonate ion (CO₃²⁻) and one hydrogen carbonate ion (HCO₃⁻).
Similarly, other reactants are represented by their respective chemical formulas.
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5.4g of aluminum reacts with sulfuric acid (H₂SO4) to form aluminum sulfate and hydrogen.
a. Write the chemical equation.
b. Find mass of required sulfuric acid.
C. Find volume of the obtained gas.
(AI=23, S = 32, O=16, H =1, 2g of H2 has 22.4L).
Answer:
a. The chemical equation for the reaction is:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
b. To find the mass of required sulfuric acid, we need to use stoichiometry. We can start by finding the number of moles of aluminum used in the reaction:
Molar mass of Al = 27 g/mol
Number of moles of Al = 5.4 g / 27 g/mol = 0.2 mol
According to the balanced equation, 3 moles of H₂SO₄ are required to react with 2 moles of Al. Therefore, the number of moles of H₂SO₄ required is:
Number of moles of H₂SO₄ = 3/2 x 0.2 mol = 0.3 mol
Molar mass of H₂SO₄ = 2 x 1 g/mol + 32 g/mol + 4 x 16 g/mol = 98 g/mol
Mass of H₂SO₄ required = 0.3 mol x 98 g/mol = 29.4 g
Therefore, 29.4 g of sulfuric acid is required to react with 5.4 g of aluminum.
c. To find the volume of hydrogen gas obtained, we need to use the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
We can start by finding the number of moles of hydrogen gas produced in the reaction. According to the balanced equation, 3 moles of H₂ are produced for every 2 moles of Al. Therefore, the number of moles of H₂ produced is:
Number of moles of H₂ = 3/2 x 0.2 mol = 0.3 mol
Assuming the reaction occurs at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol. Therefore:
V = nRT/P = 0.3 mol x 0.0821 L atm/mol K x 273 K / 1 atm = 6.58 L
Therefore, the volume of hydrogen gas produced at STP is 6.58 L.
Explanation:
Units 1 and 2 1 Choose the correct option. a) An example of a chemical change is: A melting chocolate B leaving an iron nail to rust C dissolving salt in water D allowing solid air freshener to sublimate b) Which is not a property of physical change? A Atoms and molecules are conserved. B Atoms and molecules are rearranged. C The changes are permanent. D The changes are reversible. c) Which is not a property of chemical change? A A chemical reaction takes place. B There are large changes in energy. C The products have different properties from the reactants. D They are examples of phase changes. d) Which is a wrong statement? A To prove the Law of Constant Composition you add all the atomic masses of atoms in a compound. B The Law of Constant Composition states that all samples of a chemical compound have the same elemental composition. During a chemical reaction the atoms are conserved. C D During a chemical reaction, the sum of the formula masses of the products must equal the sum of the formula masses of the reactants
A solution of KNO3 is prepared by dissolving 80 g KNO3 in 100 g water at 60oC. The solution is
Group of answer choices
dilute
saturated
unsaturated
supersaturated
A soft drink contains 33g of sugar in 349g of H2O. What is the concentration of sugar in the soft drink in mass precent?
The concentration of the sugar in the soft drink in mass percent is 8.64%
How do i determine the concentration in mass percent?First, we shall determine the mass of the solution. Details below:
Mass of sugar = 33 gramsMass of water = 349 gramsMass of solution =?Mass of solution = Mass of sugar + mass of water
Mass of solution = 33 + 349
Mass of solution = 382 grams
Finally, we shall determine the mass percent of the sugar in the solution. Details below:
Mass of sugar = 33 gramsMass of solution = 382 gramsPercentage of sugar =?Percentage = (mass of sugar / mass of solution) × 100
Percentage of sugar = (33 / 382) × 100
Percentage of sugar = 8.64%
Thus, we can conclude that the mass percent of the sugar is 8.64%
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Based on the information provided, which solution is a base and weak electrolyte
An example of a composition which fulfills the qualifications of being both a base and a weak electrolyte is ammonia (NH3).
How to explain the electrolyteA base is any constituent which voluntarily receives protons (H+) in an associated chemical reaction while an electrolyte denotes any material that can conduct electricity through liquids or in melted state.
Upon dissolution in water, it is apt to accept a proton from such and thus create the acidic ion known as ammonium (NH4+). Nonetheless, due to its scarce dissociation into hydroxide (OH-) and ammonium ions, it is deemed a weak electrolyte.
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0.250 mol of KNO3 in 0.835 L of solution
0.30M is the molarity of the given solution with 0.250 mol and 0.835 L volume of solution
The amount of moles of solute that exist in a specific number of litres of the solution, or moles per litre of a solution, is known as molar concentration or molarity. Solvent, and 'solution' to make it simpler to comprehend the principles that will follow. A homogenous combination with any number of solutes in it is referred to as a solution.
Molarity = number of moles/ volume of solution in liter
Molarity = 0.250/ 0.835
= 0.30M
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How many different genus groups are there? List them
A bag of frozen broccoli weighs 306.0 grams. You microwave it and notice a lot is steam so you weigh after microwaving and it is 275.0 grams. What happened to the percent mass of water? Show your work
There are different methods to calculate the concentration of a solution. Mass percentage is one among them. Mass percentage is mainly used to calculate the concentration of a binary solution. Here mass percent of water is 10.13.
Mass percentage of a particular component in a solution is equal to mass in grams of that component present per 100 g of the solution. For example, a 5% aqueous solution of urea means 5g of urea in 100 g of its aqueous solution.
Mass percentage = Mass of the component / Total mass of solution × 100
Mass of water = 306.0 - 275.0 = 31
% Mass = 31 / 306.0 × 100 = 10.13%
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Given the following data
C2H4 (g) + 3O2 (g) -> 2CO2 (g) + 2H2O (l) H = –1411.0 kJ
2C2H6 (g) + 7O2 (g) -> 4CO2 (g) + 6H2O (l) H = –3119.8 kJ
2H2 (g) + O2 (g) -> 2H2O (l) H = –571.7 kJ
calculate H for the reaction
C2H4 (g) + H2 -> C2H6 (g)
Answer:H = 5
Explanation: You would get this answer if you divide by 2 then multiple by 7
Calculate the oxidation number of oxygen in magnesium pyrophosphate Mg2p2o7
The oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.
o calculate the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7), we first need to know the oxidation states of the other atoms in the molecule.
The magnesium ion (Mg2+) has a fixed oxidation state of +2 in this compound, and the phosphate ion (PO43-) has an overall oxidation state of -3.
We can set up an equation to solve for the oxidation state of oxygen (O) in the pyrophosphate ion:
2(+2) + 2(O) + 7(-2) = 0
Simplifying the equation gives:
4 + 2O - 14 = 0
2O = 10
O = +5
Therefore, the oxidation state of oxygen in magnesium pyrophosphate (Mg2P2O7) is +5.
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The oxidation number of oxygen in magnesium pyrophosphate is -2.
How to calculate oxidation number?Oxidation number refers to the hypothetical charge of an atom within a molecule.
For monoatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.
Oxidation no. of Mg = +2Oxidation no. of P = +5Oxidation no of O = x0 = +2(2) + 5(2) + X(7)
0 = 14 + 7x
-14 = 7x
x = -2
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7. What mass of solid NH4Cl and what volume of 1.00 mol-L¹ NaOH solution should be used to
prepare 1 L of a buffer solution of pH 9.00? Suppose the overall concentration of the buffer is 0.125
mol-L¹. (Answer V = 45 mL)
To prepare a buffer solution of pH 9.00, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])Where [A-]/[HA] is the ratio of the concentrations of the conjugate base and acid of the buffer, respectively. Since we are given the pH and the overall concentration of the buffer, we can solve for the ratio [A-]/[HA]:
9.00 = pKa + log([A-]/[HA])pKa = 9.25 (the pKa of NH4Cl)9.00 = 9.25 + log([A-]/[HA])log([A-]/[HA]) = -0.25[A-]/[HA] = 0.56Next, we can use the definition of the concentration of a solution to find the concentration of NH4Cl needed to make a 0.125 mol-L^-1 buffer solution:
0.125 mol-L^-1 = [NH4Cl] + [NaOH]
Since the NaOH solution is 1.00 mol-L^-1, we can assume that the contribution of NaOH to the total concentration of the buffer is negligible, and so:
0.125 mol-L^-1 = [NH4Cl]Finally, we can use the molar mass of NH4Cl to find the mass of NH4Cl needed to prepare 1 L of the buffer solution:
mass NH4Cl = molar mass * molesmass NH4Cl = (14.01 g-mol^-1 + 1.01 g-mol^-1 + 35.45 g-mol^-1) * 0.125 molmass NH4Cl = 6.63 gSo we need to use 6.63 g of NH4Cl and enough volume of 1.00 mol-L^-1 NaOH solution to make a total volume of 1 L. To find the volume of NaOH solution needed, we can use the definition of molarity:
molality = moles of solute / volume of solution (in liters)Rearranging this equation, we get:
volume of solution = moles of solute / molalitySince we are adding NaOH solution to NH4Cl to make a total volume of 1 L, the molality of NaOH solution is also 0.125 mol-L^-1. Therefore:
volume of NaOH solution = moles of NaOH / molality of NaOHvolume of NaOH solution = (1 L - volume of NH4Cl solution) * 0.125 mol-L^-1Substituting the values we know:
volume of NaOH solution = (1 L - 0.45 L) * 0.125 mol-L^-1volume of NaOH solution = 0.056 L = 56 mLSo we need to use 6.63 g of NH4Cl and 56 mL of 1.00 mol-L^-1 NaOH solution to prepare 1 L of a buffer solution of pH 9.00.
NEED HELP FIGURING HOW MANY MOL!! PLEASE QUICK!!THANK YOU SO MUCH
The number of moles of the gas by the ideal gas law is 0.18 moles.
What is the ideal gas law?The behavior of an ideal gas, a hypothetical gas made up of randomly moving particles with little volume and no intermolecular interactions, is described by the ideal gas law.
Although intermolecular interactions and non-zero particle volume prevent gases from always behaving in an ideal manner, the ideal gas law is nevertheless a good approximation for many gases under some circumstances.
We know that;
PV = nRT
We have ;
P = 1.2 atm
V = 3.4 L
T = 10 + 273 = 283 K
n = ?
n = PV/RT
n = 1.2 * 3.4/0.082 * 283
n =4.08 /23.2
n = 0.18 moles
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