CHEMISTRY COLLEGE

Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answers

Answer 1

Answer:

Answer: The value of $K_c$ for the reaction is $6.005* 10^(-6)$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=(0.0150)/(1.00)=0.0150M$

$[O_2]_i=(0.0150)/(1.00)=0.0150M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial: 0.0150 0.0150

At eqllm: 0.0150-4x 0.0150-3x 2x 6x

The expression of $K_c$ for above equation follows:

$K_c=([N_2]^2[H_2O]^6)/([NH_3]^4[O_2]^3)$ .......(1)

We are given:

Equilibrium concentration of $N_2=1.96* 10^(-3)$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96* 10^(-3)\n\nx=0.98* 10^(-3)M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=((0.00196)^2* (0.00588)^6)/((0.01108)^4* (0.01206)^3)\n\nK_c=6.005* 10^(-6)$

Hence, the value of $K_c$ for the reaction is $6.005* 10^(-6)$

Answer 2

Answer:

Final answer:

To calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2, you need to determine the equilibrium concentrations of NH3 and O2. The given information includes the initial moles and concentration of NH3 and O2, as well as the equilibrium concentration of N2. Using the stoichiometry of the reaction and the given data, you can calculate the equilibrium concentrations and substitute them into the Kc expression to determine the numerical value of Kc.

Explanation:

The question asks to calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2. The reaction equation is 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g). The given information is that 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container, and the N2 concentration at equilibrium is 1.96 × 10−3 M. To solve for Kc, we need to calculate the equilibrium concentrations of NH3 and O2.

Using the stoichiometry of the reaction, we can determine that the equilibrium concentration of NH3 is (0.0150 - 2*1.96 × 10−3) M and the equilibrium concentration of O2 is (0.0150 - 3*1.96 × 10−3) M. Substituting these values into the equilibrium expression for Kc, we can calculate the value of Kc.

In this case, the equilibrium constant, Kc, can be calculated as [N2]^2 / ([NH3]^4 * [O2]^3). Substitute the given equilibrium concentration of N2 and the calculated equilibrium concentrations of NH3 and O2 into the Kc expression to determine the numerical value of Kc.

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In a 74.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.140. What is the mass of each component?

Answers

Answer:

The correct answer is 16.61 grams methanol and 57.38 grams water.

Explanation:

The mole fraction (X) of methanol can be determined by using the formula,

X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)

X₁ = n₁/n₁ + n₂ = 0.14

n₁ / n₁ + n₂ = 0.14 ---------(i)

n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)

n₁ mole CH₃OH = 32.042 n₁ g

n₂ mole H2O = n₂ mole × 18.015 g/mol

n₂ mole H2O = 18.015 n₂ g

Thus, total mole number is,

32.042 n₁ + 18.015 n₂ = 74 ------------(ii)

From equation (i)

n₁/n₁ + n₂ = 0.14

n₁ = 0.14 n₁ + 0.14 n₂

n₁ - 0.14 n₁ = 0.14 n₂

n₁ = 0.14 n₂ / 1-0.14

n₁ = 0.14 n₂/0.86 ----------(iii)

From eq (ii) and (iii) we get,

32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74

n₂ (32.042 × 0.14/0.86 + 18.015) = 74

n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)

n₂ = 3.1854 mol

From equation (iii),

n₁ = 0.14/0.86 n₂

n₁ = 0.14/0.86 × 3.1854

n₁ = 0.5185 mol

Now, presence of water in the mixture is,

= 3.1854 mole × 18.015 gram per mole

= 57.38 grams

Methanol present in the mixture is,

= 0.5185 mol × 32.042 gram per mole

= 16.61 grams

Final answer:

In a 74.0 g aqueous solution of methanol with a mole fraction of 0.140, the mass of methanol is approximately 10.36 g and the mass of water is approximately 63.64 g.

Explanation:

The problem involves the calculation of the mass of the components of an aqueous solution of methanol (CH3OH). First, we need to know that the mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the mixture.

Given that the mole fraction of methanol is 0.140, this means that the rest of the solution (i.e., water) is 1 - 0.140 = 0.860. To find the mass of each component, we need to consider the total mass of 74.0 g.

The mass of methanol can be calculated as 74.0 g * 0.140 = 10.36 g. And the mass of water would be 74.0 g * 0.860 = 63.64 g.

So, in this aqueous solution, you have approximately 10.36 g of methanol and 63.64 g of water.

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Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

Answers

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

$\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} * 100 \%$

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

$\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} * 100 \%\n\n\text{Percent Mass} = 10.6 \%}$

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

Calculate ΔHrxn for the following reaction: C(s) + H2O(g) --> CO(g) + H2(g) Use the following reactions and given ΔH values: C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ 2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ 2 H2 (g) + O2 (g) → 2 H2O (g), ΔH= -483.6 kJ

Answers

Explanation:

Below is an attachment containing the solution to the question.

The Earth can gain or lose matter. True False

Answers

Answer:

True

Explanation:

According to some calculations, the Earth is losing 50,000 metric tons of mass every single year, even though an extra 40,000 metric tons of space dust converge onto the Earth's gravity well, it's still losing weight.

Answer:

true

Explanation:

What are the names of the following compounds: FeCl HNO NaSO SO

Answers

Answer:

FeCl: Ferric Chloride (also called iron chloride), comes from Fe (ferrum, or iron), and Cl (Chlorine)

HNO: Nitroxyl, from N (Nitrogen), and the acidic nature of a radical ending in -yl.

NaSO: Sodium sulfate, Na (Sodium), S (Sulfur), O (Oxygen).

SO: Sulfur monoxide (Mono-One), O (Oxygen) and S (Sulfur).

How much energy in joules does it take to raise the temperature of 255g of water by 12.5C

Answers

Energy in a system can be calculated by multiplying the given mass to thespecific heat capacity of the substance and the temperature difference. It isexpressed as follows:

Energy = mC(T2-T1)
Energy = 255(4.184)(12.5)
Energy = 13336.5 J

Answer: The amount of energy required to raise the temperature is 13323.75 joules.

Explanation :

The amount of energy required to raise the temperature can be calculated as follows.

$q = m* C* \bigtriangleup T$

where,

q = heat energy

m = mass of water

C = specific heat

T = temperature

Remember that the specific heat of water is $4.18\left ( (J)/(g^\circ C)} \right )$ .

Therefore, putting the values in the above equation as follows.

$q = m* C* \bigtriangleup T$

$= 255 g * 4.18 \left ( (J)/(^\circ C)} \right )* 12.5^oC$

= 13323.75 joules

So, the amount of energy required to raise the temperature is 13323.75 joules.

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