a crystal is a single, continuous piece of a mineral bounded by flat surfaces that formed naturally as the mineral grew and it needs to be see-through. group of answer choices true false

When carbonates (CO3^2-) or bicarbonates (HCO3^-) are reacted with an acid in an acid-base reaction, the resulting product is carbonic acid (H2CO3).

This reaction follows the general pattern of an acid-base reaction, where the base (CO3^2- or HCO3^-) and acid (H+) combine to form the conjugate acid (H2CO3) and conjugate base (OH-).
The general equation for this reaction is:
Acid + Base ⇋ Conjugate Acid + Conjugate Base
In the case of carbonates and bicarbonates, the equation is:
H+ + CO3^2- (or HCO3^-) ⇋ H2CO3 + OH-
The reaction between carbonates and bicarbonates with an acid is called a "carbonate hydrolysis" reaction. This is because the hydroxide ions (OH-) from the reaction can hydrolyze the carbonate ion (CO3^2-) and bicarbonate ion (HCO3^-), breaking them down into carbonic acid (H2CO3).
In addition to the carbonate hydrolysis reaction, there is also a "bicarbonate hydrolysis" reaction that occurs when bicarbonate ions are reacted with an acid. The general equation for this reaction is:
H+ + HCO3^- ⇋ H2CO3 + H2O
In this reaction, the hydroxide ions are replaced with water, and the resulting product is still carbonic acid (H2CO3).

To sum up, when carbonates (CO3^2-) or bicarbonates (HCO3^-) are reacted with an acid in an acid-base reaction, the resulting product is carbonic acid (H2CO3). This reaction follows the general pattern of an acid-base reaction, where the base and acid combine to form the conjugate acid and conjugate base. The reaction between carbonates and bicarbonates with an acid is called a "carbonate hydrolysis" reaction, and for bicarbonates it is called a "bicarbonate hydrolysis" reaction.

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The first and second ionization of a diprotic acid cannot be treated as completely separate reactions when the reaction is taking place in an environment with a fixed pH.

The second ionization of the acid is dependent on the concentration of the ions produced from the first ionization.

If the pH is fixed, then the concentration of the first ionization is also fixed, so the second ionization will not occur completely independently.

For example, a diprotic acid such as oxalic acid can be completely ionized in two steps. In the first ionization, the hydrogen ions of the oxalic acid are replaced with hydroxide ions, forming the oxalate ion:

H2C2O4 + 2H2O → H3O+ + HC2O4–

In the second ionization, the oxalate ion is further dissociated, forming two separate anions and hydronium ions:

HC2O4– + H2O → H3O+ + C2O4–2

However, in an environment with a fixed pH, the second ionization will not take place as the concentration of oxalate ions from the first ionization is fixed.

Therefore, the two ionizations must be treated together in order to accurately predict the final concentrations of the products.

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Answer:

The first ionization constant is greater than the second ionization constant by only a factor of 10.

Explanation:

The two ionization constants must differ by a factor of at least 20 in order to treat the first and second ionizations as chemically (and mathematically) distinct.

Answer:

Na (s) + Cl2 (g) → NaCl (s)

Explanation:

A reaction of sodium with chlorine to produce sodium chloride is an example of a combination reaction. 2 Na + Cl 2 → 2 NaCl.

Universal waste is a category of hazardous waste that includes certain widely generated electronic devices, batteries, lamps, and other devices that contain hazardous materials.

The handling, storage, transportation, and disposal of universal waste is subject to regulations by the United States Environmental Protection Agency (EPA) under the Universal Waste Rule.

One of the requirements of the Universal Waste Rule is that records of universal waste shipments must be retained for a minimum of three years. This applies to any person who generates, collects, transports, or receives universal waste. The records must include the following information:

Retention of these records helps to ensure compliance with the regulations and enables tracking of the movement and disposition of universal waste. The records must be made available for inspection by authorized EPA officials upon request.

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Answer: The amount of heat energy required to melt 649.2 g of HBr is 12.99 kJ, given that the molar heat of fusion of HBr is 2.41 kJ/mol.

Molar heat of fusion is the amount of heat required to melt one mole of a substance. The molar heat of fusion for HBr is 2.41 kJ/mol.

To find the amount of heat energy required to melt 649.2 g of HBr, the following steps should be followed:

Step 1: Determine the number of moles of HBr in 649.2 g of HBr:mass of HBr = 649.2 gMolar mass of HBr = 80.91 g/molNumber of moles of HBr = mass/molar mass= 649.2 g/80.91 g/mol= 8.01 mol

Step 2: Calculate the amount of heat required to melt 1 mol of HBr:Given molar heat of fusion of HBr is 2.41 kJ/molHeat required to melt 1 mol of HBr = 2.41 kJ/mol

Step 3: Calculate the amount of heat required to melt 8.01 mol of HBr:Heat required to melt 8.01 mol of HBr = Heat required to melt 1 mol of HBr × Number of moles of HBrHeat required to melt 8.01 mol of HBr = 2.41 kJ/mol × 8.01 molHeat required to melt 8.01 mol of HBr = 19.301 kJ

Step 4: Convert the heat in kJ to J by multiplying it with 1000: Heat required to melt 8.01 mol of HBr = 19.301 kJ = 19,301J. Finally, we get the result: The amount of heat energy required to melt 649.2 g of HBr is 12.99 kJ.

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In this case, if the reaction produces ketones, the infrared spectrum should show peaks associated with the C=O and C-H bonds of the ketones, but no peaks associated with the starting material.

The infrared spectrum of a reaction can be used to identify the starting material and products in a reaction. If a reaction is complete, there should be no peaks associated with the starting material, only the products. There are two ways to determine the absence of the starting material, and these are as follows:

As a result, if that peak or band is absent, it is evident that the starting material has been entirely consumed. To demonstrate that the products are ketones, there are several bands present in the IR spectrum, which can be looked for, and these are as follows:

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The IR spectrum can be used to identify ketones due to the presence of a strong C=O bond, which results in a characteristic absorption peak around 1730 cm-1. A comparison of the IR spectrum of the starting material and product can be used to confirm that the starting material is completely consumed and the products are ketones.

To demonstrate that there is no beginning material left and that the products are ketones, the IR spectrum can be used. Infrared (IR) spectroscopy is a technique that measures the absorbance of infrared radiation in a substance. When a compound absorbs infrared light, it vibrates at a particular frequency, which is dependent on the chemical structure of the compound. By studying these vibrational frequencies, the IR spectrum of a sample can reveal a great deal about its molecular structure and composition.

IR spectroscopy can be used to show that the starting material has been fully consumed and that the products are ketones. During a reaction that transforms a ketone from a different compound, the IR spectrum of the product will exhibit a carbonyl (C=O) peak at around 1710 cm-1. The absence of peaks corresponding to the beginning material in the product's IR spectrum indicates that the beginning material has been completely consumed. If a new peak that corresponds to the C=O bond appears in the IR spectrum of the product, this shows that the reaction has produced a ketone.

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The number of mole present in 87 grams of potassium bromide, KBr is 0.731 mole

We'll begin our calculation by obtaining the molar mass of potassium bromide, KBr. Details below:

Molar mass of potassium bromide, KBr = K + Br

Molar mass of potassium bromide, KBr = 39 + 80

Molar mass of potassium bromide, KBr = 119 g/ mol

Finally, we shall determine the number of mole present. Details below:

Mole = mass / molar mass

Mole of potassium bromide, KBr = 87/ 119

Mole of potassium bromide, KBr = 0.731 mole

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Complete question:

How many moles are in 87g of potassium bromide?

There are 1.49 × 10²³ formula units of potassium nitrate in a 25 g sample.

One formula unit is defined as the simplest formula of a substance, which indicates the relative amounts of the elements in the molecule. As a result, the number of formula units in a sample can be calculated by dividing the sample's mass by the substance's molar mass.

The molecular formula of potassium nitrate is KNO3. It contains one potassium atom (K), one nitrogen atom (N), and three oxygen atoms (O). The atomic masses of the elements can be used to calculate the molar mass of the compound.

One potassium atom has a molar mass of 39.1 g/mol, one nitrogen atom has a molar mass of 14.0 g/mol, and three oxygen atoms have a combined molar mass of 48.0 g/mol.

The molar mass of KNO3 = (1 × 39.1 g/mol) + (1 × 14.0 g/mol) + (3 × 16.0 g/mol) = 101.1 g/mol.

Now, on dividing the sample's mass (25 g) by the molar mass of potassium nitrate (101.1 g/mol), a value of 0.247 mol is obtained. The Avogadro constant can be used to convert moles into formula units. The Avogadro constant, 6.022 × 10²³ formula units per mole, represents the number of formula units in one mole of a substance.

The number of formula units = (0.247 mol) × (6.022 × 10²³ formula units/mol) = 1.49 × 10²³ formula units.

Therefore, there are 1.49 × 10²³ formula units of potassium nitrate in a 25 g sample.

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In the chemical equations, the reactants can be classified as follows:

1. O2 is an oxidizing agent as it gains electrons and gets reduced.
2. MnO2 is an oxidizing agent as it gains electrons and gets reduced.
3. ZnS is a reducing agent as it loses electrons and gets oxidized.
4. Cu2+ is an oxidizing agent as it gains electrons and gets reduced.
5. H2S is a reducing agent as it loses electrons and gets oxidized.
6. Cl- is a reducing agent as it loses electrons and gets oxidized.
7. H+ is an oxidizing agent as it gains electrons and gets reduced.

Now, let's calculate the increase or decrease in the oxidation state for each element as it changes from a reactant to a product:

1. Sulfur, beginning in the reactant ZnS, has an oxidation state of -2. In the product SO2, sulfur has an oxidation state of +4. The change in oxidation state is +4 - (-2) = +6.

2. Sulfur, beginning in the reactant H2S, has an oxidation state of -2. In the product CuS, sulfur has an oxidation state of -2. The change in oxidation state is -2 - (-2) = 0.

3. Chlorine, beginning in the reactant Cl-, has an oxidation state of -1. In the product Cl2, chlorine has an oxidation state of 0. The change in oxidation state is 0 - (-1) = +1.

4. Manganese, beginning in the reactant MnO2, has an oxidation state of +4. In the product Mn2+, manganese has an oxidation state of +2. The change in oxidation state is +2 - (+4) = -2.

So the oxidation state changes are:
Sulfur in ZnS = +6
Sulfur in H2S = 0
Chlorine in Cl- = +1
Manganese in MnO2 = -2

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Evaporites form as ions (minerals) precipitate out of an evaporating solution. The correct option is B.

Evaporites are minerals that are created as a result of the evaporation of water. The minerals are usually found in salt pans or salt lakes. Salt pans are shallow pans that are usually found in hot and dry regions of the world. In most cases, salt pans are usually found in places where water sources are limited. Evaporites form as ions (minerals) precipitate out of an evaporating solution.

As the water evaporates, it leaves behind salt crystals. Over time, these salt crystals can build up and form a layer of salt. The process of evaporation and deposition can repeat itself many times over the years, resulting in the formation of thick layers of salt.

There are different types of evaporites, and they are classified based on the minerals that are formed. Some of the most common types of evaporites include halite, gypsum, and anhydrite. Halite is the most common type of evaporite, and it is usually found in salt pans and salt lakes. Gypsum and anhydrite are usually found in areas that have been submerged in water for long periods of time.

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Chlorofluorocarbon (CFC) is not an example of a naturally occurring greenhouse gas.

CFCs are human-made gases that are not naturally found in the atmosphere. These gases trap heat in the atmosphere, contributing to the greenhouse effect, but are not naturally produced.

On the other hand, methane, nitrous oxide, and water vapor are all naturally occurring greenhouse gases.

Methane is produced by microbial processes in the environment, while nitrous oxide and water vapor come from naturally occurring processes like volcanoes and evaporation.

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Answer:

Explanation:

I think 'The heat from Earth's core causes material in the area under the crust to become denser and rinse, while less dense material sinks.

The combination that will produce an insoluble salt is b) MnSO4 Pb(NO2)2.

A salt is a chemical compound made up of cations (positively charged ions) and anions (negatively charged ions) (negatively charged ions). The ions must be combined in such a way that the sum of the charges is zero. NaCl is the most well-known saltand it is made up of sodium cations (Na+) and chloride anions (Cl-).MnSO4 Pb(NO2)2 is the answer since both of these elements are soluble. MnSO4 is a soluble substance that is sometimes used in the production of ceramics.

MnSO4 is often used as a nutritional supplement for animals since it is a good source of manganese. Pb(NO2)2 is a powder that is bright yellow, it has a molar mass of 325.2 g/mol. It is made up of two NO2 anions (negatively charged ions) and one Pb2+ cation (positively charged ion).The formation of insoluble salts can occur when the cations and anions in a reaction solution bind to create a new solid. Since the newly formed solid is insoluble, it settles to the bottom of the solution and can be separated from the liquid through filtration. The insoluble salt that is formed is a white or colorless substance that appears as a powder.

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Answer:

74g.

Explanation:

The volume won't increase by the volume of salt you added, though. This is for many different reasons among them the fact that salt is in grains (with lots of air in between) and the salt dissolving in the water and kind of squeezing in the spaces between water molecules. But the mass should increase by exactly the 19g you added.

The ground state electron configuration of an atom of the element identified in the mass spectrometer results is  1s²2s²2p⁶3s².

The sample of the pure element that is analyzed using a mass spectrometer shows the following results:

Bar one: amu 24 and percent abundance just below 80.

Bar 2: amu 25 and percent abundance 10

Bar 3: amu 26 and percent abundance just above 10.

The ground-state electron configuration of an atom of the element that is identified in part c is as follows:

The mass number of the element is the weighted average of the isotopic masses, and it is calculated by adding the product of each isotope's atomic mass and its percent abundance. The calculation for the above-given values is shown below:

(24 amu × 0.79) + (25 amu × 0.10) + (26 amu × 0.11) = 24.33 amu

Since the mass number of the element is closer to 24 than to 25, it is reasonable to believe that the element is magnesium (Mg). The atomic number of magnesium is 12. Therefore, its electron configuration in the ground state is 1s²2s²2p⁶3s².

Hence, the ground-state electron configuration of an atom of the element that you identified in part c is 1s²2s²2p⁶3s².

Complete answer:

A sample of a pure element is anylazed using a mass spectrometer. The results are shown below.

Bar one: amu 24 and percent abundance just below 80.

Bar 2: amu 25 and percent abundance 10

Bar 3: amu 26 and percent abundance just above 10.

Write the ground-state electron configuration of an atom of the element that you identified in part c.

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As a result, oxygen exerts a pressure of 0.5 atm.

1013.25 mbar is the atmospheric pressure at sea level (under normal atmospheric circumstances). Here, nitrogen (78.08% vol), oxygen (20.95% vol), argon (0.93% vol), and carbon dioxide (0.040% vol) make up the majority of the dry air.

If nitrogen is responsible for 90% of the total pressure, oxygen is responsible for the remaining 10%.

First, let's calculate the pressure that nitrogen exerts:

Pressure of nitrogen = 90% of total pressure

= 0.9 * 5.0 atm

= 4.5 atm

Now, we can find the pressure exerted by oxygen:

Pressure of oxygen = 10% of total pressure

= 0.1 * 5.0 atm

= 0.5 atm.

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a. The parentheses indicate that the elements within are grouped together, i.e., they are part of the same unit.

Tribarium Phosphate,Tribarium is the chemical symbol for Barium and the name reflects the fact that it is composed of three atoms of Barium and two atoms of Phosphate.It is an inorganic salt that is insoluble in water and has a variety of uses in industrial and medical applications.

b. The subscript 4 indicates that there are four [tex]PO_4\ molecules[/tex] in the compound.

c.The subscript 2 implies that there are two [tex]Ba_3[/tex] molecules in the compound.

d. The coefficient indicates the number of molecules of each element in the compound. In this case, there is one [tex]Ba_3[/tex] molecule, four [tex]PO_4[/tex] molecules, and three Ca molecules.

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Yes, a solid Cu(OH)2 will form when 0.075 g KOH is dissolved in 1.0 L of 1.0 x 10^-3 M Cu(NO3)2.  0.107 g of solid Cu(OH)2 will form.

First, we need to determine the amount of Cu2+ ions present in the solution:
1.0 x 10^-3 M Cu(NO3)2 means that there are 1.0 x 10^-3 moles of Cu2+ ions per liter of solution.
Next, we can use stoichiometry to determine the amount of OH- ions that will react with the Cu2+ ions to form Cu(OH)2. The balanced chemical equation for this reaction is:
Cu2+ (aq) + 2OH- (aq) → Cu(OH)2 (s)
For every 1 mole of Cu2+ ions, we need 2 moles of OH- ions. Therefore, the total amount of OH- ions needed to react with all of the Cu2+ ions in the solution is:
2 x 1.0 x 10^-3 mol = 2.0 x 10^-3 mol
Now we can use the Ksp of Cu(OH)2 to calculate the concentration of Cu2+ and OH- ions in the solution. The Ksp expression for Cu(OH)2 is:
Ksp = [Cu2+][OH-]^2
Since we know the Ksp value for Cu(OH)2, we can solve for either [Cu2+] or [OH-]. Let's solve for [OH-]:
Ksp = [Cu2+][OH-]^2
4.8 x 10^-20 = (1.0 x 10^-3 M)[OH-]^2
[OH-]^2 = 4.8 x 10^-17
[OH-] = 2.2 x 10^-9 M
Therefore, the concentration of OH- ions in the solution is 2.2 x 10^-9 M. Since we need 2 moles of OH- ions for every mole of Cu2+ ions, we know that the concentration of Cu2+ ions is half of the concentration of OH- ions:
[Cu2+] = 1.1 x 10^-9 M
Finally, we can use the molar mass of Cu(OH)2 to determine the mass of solid that will form:
Molar mass of Cu(OH)2 = 97.56 g/mol
1 mole of Cu(OH)2 is formed for every mole of Cu2+ ions, so the mass of Cu(OH)2 that will form is:
0.0011 mol x 97.56 g/mol = 0.107 g
Therefore, 0.107 g of solid Cu(OH)2 will form when 0.075 g KOH is dissolved in 1.0 L of 1.0 x 10^-3 M Cu(NO3)2.

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When looking down the C2-C3 bond of pentane, the staggered conformations have the same representation (show the same orientation) there are three staggered conformations

Isomers are molecules with the same formula but a different spatial orientation of the atoms, meaning they have different shapes. Conformations refer to the different spatial arrangements that a molecule can take on by rotating around single bonds, such as those in pentane. The staggered conformations, which occur when the two largest substituents are 60 degrees apart, are the most thermodynamically stable of the conformations for pentane.

Therefore, when looking down the C2-C3 bond of pentane, there are three staggered conformations that have the same representation (show the same orientation).

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By plugging in the values for each of the parameters and solving for t, the time required for 99.9% of the 2-chloro-2-methylpropane to hydrolyze can be determined.

The time required for 99.9% of the 2-chloro-2-methylpropane to hydrolyze at room temperature depends on the specific conditions of the reaction. Generally, it will take several hours for this reaction to occur.

To calculate the exact time required, we can use the Arrhenius equation, which is given as:

   k = A*e(-Ea/RT)

Where:

   k = rate constant for the reaction

   A = pre-exponential factor

   Ea = activation energy

   R = gas constant

   T = temperature

The values for each of the parameters and solving for t in the equation, the time required for 99.9% of the 2-chloro-2-methylpropane to hydrolyze can be determined.

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The samples that  has the most moles of the compound is option B which is 75.0g

To determine which sample has the most moles of the compound, we need to calculate the number of moles of each compound using its molar mass.

a) Fe2O3:

Molar mass of Fe2O3 = 2(55.85 g/mol of Fe) + 3(16.00 g/mol of O) = 159.70 g/mol

Number of moles of Fe2O3 = 163.0 g / 159.70 g/mol = 1.02 mol

b) CaS:

Molar mass of CaS = 40.08 g/mol of Ca + 32.06 g/mol of S = 72.14 g/mol

Number of moles of CaS = 75.0 g / 72.14 g/mol = 1.04 mol

Therefore, sample b) (75.0 g of CaS) has the most moles of the compound, with 1.04 moles. Sample a) (163.0 g of Fe2O3) has 1.02 moles and sample c) (150.0 g of BaO) has 0.98 moles.

So, the correct answer is b.

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As temperature increases, vapor pressure of substance also increases due to an increase in  kinetic energy of the molecules. The correct answers are options: 1, 2, 3, 4.

As temperature increases, vapor pressure of a substance also increases due to an increase in  kinetic energy of molecules Substances with stronger intermolecular forces will have lower vapor pressure because it requires more energy to break bonds between molecules and transition into  gas phase. An increase in external pressure will decrease  vapor pressure. Molecular size and shape of a substance can affect intermolecular forces and therefore its vapor pressure. For example, larger molecules tend to have stronger intermolecular forces, which result in lower vapor pressures. Options are 1, 2, 3, 4  correct .

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--The complete Question is, which of the following will affect the vapor pressure of a pure molecular substance?

select all that apply.

1. the external pressure

2. the structure of the substance

3. the strength of the intermolecular forces

4. the temperature

5. the weather conditions--

If 0.0200 m fe3 is initially mixed with 1.00 m oxalate ion, then concentration of Fe3+ ion at equilibrium is 0 M.

The balanced chemical equation for the reaction of Fe3+ ion and oxalate ion is:

Fe3+ + 3C2O42- -> Fe(C2O4)33-

The reaction quotient, Qc, for the above reaction is given by the expression:

Qc = [Fe(C2O4)33-]/[Fe3+][C2O42-]

Here, the initial concentration of Fe3+ ion

= 0.0200 m

And, the initial concentration of oxalate ion is 1.00 m . According to the stoichiometry of the balanced equation, 1 mole of Fe3+ ion reacts with 3 moles of C2O42- ions to form 1 mole of Fe(C2O4)33- complex ion. Hence, the concentration of C2O42- ion that reacts with the given initial concentration of Fe3+ ion is given by the expression: [C2O42-] = 3[Fe3+] = 3 x 0.0200 m = 0.0600 m. After the reaction comes to equilibrium, let the concentration of Fe3+ ion be x M.Now, [Fe(C2O4)33-] = 0 M (as the entire Fe3+ ion is converted into Fe(C2O4)33- complex ion)Substituting the given and calculated values in the expression for Qc, we get:

Kc = [Fe(C2O4)33-]/[Fe3+][C2O42-]

=> 0/[x][0.0600]

=> 0x

=> 0 M

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The volume of 0.150 M HNO3 that can be prepared from 0.350 L of 1.98 M HNO3 is 0.07112 L, or approximately 71.12 mL (since 1 L = 1000 mL).

The given equation is used to calculate the volume (V1) of a desired concentration of a solution (0.150 M HNO3) that can be prepared from a given volume (V2) of a known concentration solution (1.98 M HNO3), using the ratios of their concentrations (C1 and C2).

Let's break down the calculation step by step using the given values:

V2 (given volume) = 0.350 L

C1 (desired concentration) = 0.150 M

C2 (known concentration) = 1.98 M

Plugging these values into the equation, we get:

V1 (0.150 M HNO3) = V2 (1.98 M HNO3) x (C1 (0.150 M) / C2 (1.98 M))

V1 = 0.350 L x (0.150 M / 1.98 M)

V1 = 0.350 L x 0.0758

V1 = 0.07112 L

Therefore, the volume of 0.150 M HNO3 that can be prepared from 0.350 L of 1.98 M HNO3 is 0.07112 L, or approximately 71.12 mL (since 1 L = 1000 mL).

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The excess reagent is FeCl₃ and 0.33 moles of FeCl₃ remains unreacted after igniting 3.00 moles of FeCl₃ with 2.00 moles of O₂ gas.

The balanced chemical equation for the reaction between FeCl₃ and O₂ is:

4 FeCl₃ + 3 O₂ → 2 Fe₂O₃ + 6 Cl₂

From the balanced equation, we can see that for every 4 moles of FeCl₃, we need 3 moles of O₂.

To determine what is the excess reagent and how much of it is left over, we need to calculate the amount of each reagent required for complete reaction:

3.00 moles FeCl₃ × (3 moles O₂/4 moles FeCl₃) = 2.25 moles O₂ required

2.00 moles O₂ × (4 moles FeCl₃/3 moles O₂) = 2.67 moles FeCl₃ required

Since we only have 2.00 moles of O₂ available, this is the limiting reagent and there is not enough O₂ to react completely with all 3.00 moles of FeCl₃. Therefore, FeCl₃ is the excess reagent.

The amount of excess reagent remaining can be calculated by subtracting the amount required for complete reaction from the amount initially present:

Excess FeCl₃ = 3.00 moles - 2.67 moles = 0.33 moles

Therefore, there is an excess of 0.33 moles of FeCl₃ remaining unreacted. There is no excess of O₂ remaining, as we started with less than the amount required for complete reaction.

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The equilibrium constant for the new reaction at 752 K is approximately 0.29 and at 719 K is approximately 0.42.

Step wise explanation:

1) For the first reaction, the equilibrium constant (Kc) is given as 11.8 at 752 K for the reaction:

[tex]2NH_{3}[/tex](g) ⇌ [tex]N_{2}[/tex](g) + [tex]3H_{2}[/tex](g)

You are asked to calculate Kc for the following reaction:

[tex]1/2N_{2} + 3/2H_{2}[/tex] ⇌ [tex]NH_{3}[/tex](g)

To find Kc for the new reaction, note that it is the reverse of the original reaction with all coefficients divided by 2. To calculate the equilibrium constant for the reverse reaction, take the reciprocal of the original Kc, and then raise it to the power of the coefficients ratio (1/2):

Kc (new) =[tex]\sqrt{ (1 / Kc (original))}[/tex] = [tex]\sqrt{(1 / 11.8)}[/tex] ≈ 0.29

So, the equilibrium constant for the new reaction at 752 K is approximately 0.29.

2) For the second reaction, the equilibrium constant (Kc) is given as 5.70 at 719 K for the reaction:

[tex]2NH_{3}[/tex](g) ⇌ [tex]N_{2}[/tex](g) + [tex]3H_{2}[/tex](g)

You are asked to calculate Kc for the following reaction:

[tex]NH_{3}[/tex](g) ⇌ [tex]1/2N_{2}[/tex](g) + [tex]3/2H_{2}[/tex](g)

This new reaction is the reverse of the original reaction with all coefficients divided by 2. Similar to the first case, take the reciprocal of the original Kc and then raise it to the power of the coefficients ratio (1/2):

Kc (new) = [tex]\sqrt{(1 / Kc (original))}[/tex] = [tex]\sqrt{(1 / 5.70)}[/tex] ≈ 0.42

So, the equilibrium constant for the new reaction at 719 K is approximately 0.42.

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Answer:

0.6096

Explanation:

*formula for moles= mass/molormass(RFM)

Molarmass= (28×1)+(19×4)= 104

63.4/104= 0.60961

The molar mass of the solute x is 85.32 g/mol.

The mass of the unknown, non-volatile, non-electrolyte solute = 12.0 g

Mass of the solvent = 100 g

The vapor pressure of the solvent before adding the solute = 100 torr

The vapor pressure of the solvent after adding the solute = 91.4 torr

Temperature = 299 K

Raoult's law can be written as:

P₂ = X₂ * P₁

Where:

P₁ = the vapor pressure of the pure solvent

P₂ = the vapor pressure of the solution

X₂ = the mole fraction of the solute

Solving for

X₂;X₂ = P₂/P₁ = 91.4/100

    X₂ = 0.914

Calculate the moles of benzene;

n = 100g / 78.11 g/mol = 1.28 moles

X₂ = moles of solute / (moles of solute + moles of benzene)

Substituting the value of X₂ and moles of benzene;

n = 0.1406 moles

Now we need to calculate the moles of the solute;

Mass of solute = 12.0 g

Now, we will use the following formula to calculate the molar mass of the solute;

Molar mass = Mass of solute / Moles of solute

Molar mass = 12.0 g / 0.1406 moles

Molar mass of the solute is 85.32 g/mol.

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An oxide of rhenium crystallizes with rhenium atoms at each of the corners and oxygen atoms in the middle of each edge of the cell. The formula of the oxide is ReO2.

An oxide is a chemical compound of at least one oxygen atom and one other element. One of the most common oxides is carbon dioxide, which is made up of one carbon atom and two oxygen atoms. Oxides are found in many other minerals and rocks, as well as in the atmosphere and they may be divided into acidic oxides and basic oxides on the basis of their chemical behavior. Formula of the oxide the rhenium atoms are present at each of the corners, whereas the oxygen atoms are located in the middle of each edge of the cell, as a result, the oxide formula will be ReO2.

When the structure of the oxide is observed, it is observed that the oxide is made up of tetrahedra in which the oxygen atoms are positioned at the vertices and the rhenium atoms are positioned at the centre of each face. In this, the atoms are arranged in the form of a cubic unit cell, with the oxygen atoms situated at each corner and the rhenium atoms located at the center of each edge. As a result, the oxide formula will be ReO2.

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The pH of hydroxylamine will be 8.76.

The first step is to write the balanced equation for the reaction of hydroxylamine with water:

NH₂OH + H₂O ⇌ NH₃OH⁺ + OH⁻

The Kb expression for this reaction is:

Kb = [NH₃OH⁺][OH⁻] / [NH₂OH]

We are given the Kb value as 6.6 x 10⁻⁹, so we can use this to find the concentration of hydroxylamine that has been deprotonated:

Kb = [NH₃OH⁺][OH⁻] / [NH₂OH]

6.6 x 10⁻⁹ = x² / (0.050 - x)

Assuming that x is very small compared to 0.050, we can simplify the expression as follows:

6.6 x 10⁻⁹ = x² / 0.050

x² = 3.3 x 10⁻¹⁰

x = 5.7 x 10⁻⁶ M

Now that we have the concentration of hydroxide ions, we can use this to find the pH of the solution:

pOH = -log[OH-] = -log(5.7 x 10⁻⁶) = 5.24

pH = 14.00 - pOH = 8.76

Therefore, the pH of a 0.050 M solution of hydroxylamine is 8.76.

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