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3 years ago

Write the balanced chemical equation for the reaction between aqueous sodium carbonate and

Chemistry

1 answer:

ioda3 years ago

7 0

Answer:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

Explanation:

This question is asking to write and balance an equation between between aqueous sodium carbonate (Na2CO3) and aqueous nitric acid (HNO3). The equation is as follows:

HNO3 (aq) + Na2CO3 (aq) → NaNO3 (aq) + CO2 (g) + H2O (l)

However, this equation is not balanced as the number of atoms of each element must be the same on both sides of the equation. To balance the equation, one will make use of coefficients as follows:

2HNO3 (aq) + Na2CO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O (l)

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At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (

Sveta_85 [38]

Answer: The equilibrium constant for the above reaction is 1.31

Explanation:

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 0.3411 1.661

At eqllm: 0.3411-x 1.661-3x 2x

Evaluating for 'x', we get:

$\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141$

Volume of the container = 2.50 L

The expression of $K_c$ for the above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

We are given:

$[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M$

$[N_2]=\frac{0.2001}{2.5}=0.08004M$

$[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M$

Putting values in above expression, we get:

$K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31$

Hence, the equilibrium constant for the above reaction is 1.31

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3 years ago

The specific heat of water is 4.18 J/(g⋅∘ c. Calculate the molar heat capacity of water.

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We are given that the specific heat of water is 4.18 J / g °C. We know that the molar mass of water is 18.02 g/mol, therefore the molar heat capacity is:

molar heat capacity = (4.18 J / g °C) * 18.02 g / mol

molar heat capacity = 75.32 J / mol °C

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anastassius [24]

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