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Sphinxa [80]
2 years ago
11

Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )

Chemistry
1 answer:
Olenka [21]2 years ago
3 0

Answer:

\large \boxed{\text{-486 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                      2H₂   +   O₂ ⟶ 2H-O-H

Bonds:          2H-H    1O=O       4H-O

D/kJ·mol⁻¹:     436      498          464

\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.

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Which of the following actions would cause the volume of the chamber below to increase to 36.0 L?
Orlov [11]

Answer:

Option C. Triple the number of moles

Explanation:

From the ideal gas equation:

PV = nRT

Where:

P is the pressure

V is the volume

n is the number of mole

R is the gas constant

T is the absolute temperature.

Making V the subject of the above equation, we have:

PV = nRT

Divide both side by P

V = nRT / P

Thus, we can say that the volume (V) is directly proportional to both the number of mole (n) and absolute temperature (T) and inversely proportional to the pressure (P). This implies that and increase in either the number of mole, the absolute temperature and a decrease in the presence will cause the volume to increase.

Thus, the correct option is option C triple the number of moles. This can further be seen as illustrated below:

Initial volume (V1) = 12 L

Initial mole (n1) = 0.5 mole

Final mole (n2) = triple the initial mole = 3 × 0.5 = 1.5 mole

Final volume (V2) =?

From:

V = nRT / P, keeping T and P constant, we have:

V1/n1 = V2/n2

12/0.5 = V2/1.5

24 = V2/1.5

Cross multiply

V2 = 24 × 1.5

V2 = 36 L.

Thus Option C gives the correct answer to the question.

5 0
3 years ago
What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)
Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\  ml\\\\V = 33.33 \ ml

Therefore, volume required is 33.33 ml .

5 0
3 years ago
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tatyana61 [14]

Answer:

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Explanation:

Given data:

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Initial volume = 3.00 L

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Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ /  P₂

V₂ =  5.00 atm × 3.00 L / 1 atm

V₂ =  15.00 atm

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