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olchik [2.2K]
2 years ago
6

In the presence of excess oxygen, methane gas burns in a constant-pressure system to yield carbon dioxide and water: CH4 (g) + 2

O2 (g) → CO2 (g) + 2H2O (l) ΔH = -890.0 kJ Calculate the value of q (kJ) in this exothermic reaction when 1.80 g of methane is combusted at constant pressure.
Chemistry
1 answer:
Gwar [14]2 years ago
8 0

<u>Answer:</u> The value of q for the reaction will be -100.1 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of methane = 1.80 g

Molar mass of methane = 16 g/mol

Putting values in above equation, we get:

\text{Moles of }CH_4=\frac{1.80g}{16g/mol}=0.1125mol

For the given chemical reaction:

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H^o_{rxn}=-890.0kJ

By Stoichiometry of the reaction:

When 1 mole of methane reacts, the heat released is 890.0 kJ

So, when 0.1125 moles of methane will react, the heat released will be \frac{890.0kJ}{1mol}\times 0.1125mol=100.1kJ

<u>Sign convention of heat (q):</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

Hence, the value of q for the reaction will be -100.1 kJ

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This combination will form a buffer.

Explanation

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Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

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Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

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Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

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Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

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Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

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The answer:

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When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
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Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

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Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

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The new molar concentration of CO at equilibrium will be:

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Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

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Best regards.

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Data obtained from the question include:

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