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2 years ago

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

Chemistry

1 answer:

posledela2 years ago

5 0

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)$

The expression for $K_c$ is written as:

$K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}$

$K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}$

$K_c=3.72$

Thus the value of the equilibrium constant Kc for this reaction is 3.72

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A student dissolved 3.50g of copper (II) nitrate in water and mixed it with a solution of sodium carbonate. The student recovere

Arisa [49]

Answer: If the theoretical yield is 2.35g then the percent yield is 80.4%

Explanation:

Given: Actual yield = 1.89 g

Theoretical yield = 2.35 g

Formula used to calculate the percentage yield is as follows.

$Percent yield = \frac{actual yield}{theoretical yield} \times 100$

Substitute the values into above formula as follows.

$Percent yield = \frac{actual yield}{theoretical yield} \times 100\\= \frac{1.89}{2.35} \times 100\\= 80.4 percent$

Thus, we can conclude that if the theoretical yield is 2.35g then the percent yield is 80.4%

3 0

2 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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2 years ago

A certain system absorbs 350 joules of heat and has 230 joules of work done on it what is the value of

marysya [2.9K]

According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt

(10.0 g) / (0.0480 mol) = 208.3 g/mol

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71

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2 years ago

Write formula of compound given two chemicals

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2 years ago

Or a hydrogen atom, which electronic transition would result in the emission of a photon with the highest energy?

ELEN [110]

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8 0

2 years ago