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astra-53 [7]
3 years ago
8

At 25°C and constant pressure, carbon monoxide gas combines with oxygen gas to give carbon dioxide gas with the evolution of 10.

1 kJ per gram of carbon monoxide consumed. What is the value of ΔH for the reaction as represented by the equation below.
Chemistry
1 answer:
stealth61 [152]3 years ago
4 0

Answer : The value of \Delta H for the reaction is, -565.6 kJ

Explanation :

First we have to calculate the molar mass of CO.

Molar mass CO = Atomic mass of C + Atomic mass of O  = 12 + 16 = 28 g/mole

Now we have to calculate the moles of CO.

\text{Moles of }CO=\frac{\text{Mass of }CO}{\text{Molar mass of }CO}=\frac{1g}{28g/mole}=\frac{1}{28}mole

Now we have to calculate the value of \Delta H for the reaction.

The balanced equation will be,

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

From the balanced chemical reaction we conclude that,

As, \frac{1}{28}mole of CO release heat = 10.1 kJ

So, 2 mole of CO release heat = 2\times 28\times 10.1=565.6kJ

Therefore, the value of \Delta H for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)

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Answer:

P_{N_2}=1.0atm

Explanation:

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Considering the Dalton's law which states that the total pressure of a gaseous system is defined by the summation of the the partial pressures of the present gases:

P_T=\Sigma P_i

For the given system:

P_T=P_{H_2}+P_{N_2}+P_{CH_4}

Solving for the partial pressure of nitrogen we obtain:

P_{N_2}=3.0atm-0.5atm-1.5atm=1.0atm

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3 years ago
Write the structure of the aldol condensation-dehydration product that you synthesized. Using this structure, analyze the NMR sp
STALIN [3.7K]

Answer:

Please find the solution in the attached file.

Explanation:

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2 years ago
How does a system at equilibrium respond to the addition of more reactant
grin007 [14]

Answer:

It favors the forward reaction.

Explanation:

According to Le Chatelier's Principle, when a system at equilibrium suffers a perturbation, the system will react in order to counteract the effect of such perturbation.

If more reactant is added, the system will try to decrease its concentration. It will do so by favoring the forward reaction, decreasing the concentration of the reactant  and increasing the concentration of the products, in order to re-establish the equilibrium.

7 0
3 years ago
The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at
kherson [118]

<u>Answer:</u> The molecular formula of the compound is C_4H_{10}O_4

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol

  • <u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=36.26g

Mass of H_2O=14.85g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, \frac{12}{44}\times 36.26=9.89g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, \frac{2}{18}\times 14.85=1.65g of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = \frac{0.824}{0.659}=1.25\approx 1

For Hydrogen = \frac{1.65}{0.659}=2.5

For Oxygen = \frac{0.659}{0.659}=1

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is C_2H_5O_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

n=\frac{133.9g/mol}{61g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4

Hence, the molecular formula of the compound is C_4H_{10}O_4

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