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Gala2k [10]
3 years ago
11

Calculate the approximate enthalpy change, δhrxn, for the combustion of one mole of methane a shown in the balanced chemical equ

ation: ch4+2o2→2h2o+co2 use the values you calculated in parts a, b, c, and d, keeping in mind the stoichiometric coefficients
Chemistry
1 answer:
Elena L [17]3 years ago
4 0
To solve for the enthalpy of reaction, we apply the Hess's Law.

ΔHrxn = ∑(ν×Hf of products) - ∑(ν×Hf of reactants)
where
v is the stoichiometric coefficient determined from the balanced reaction
Hf is the standard heat of formation; these are empirical values:
*For CH₄: Hf = <span>−74.87 kJ/mol
*For O</span>₂: Hf = 0
*For CO₂: <span>-393.5 kJ/mol 
*For H</span>₂O: <span>-241.82 kJ/mol

</span>ΔHrxn = [(2*-241.82 kJ/mol)+(1*-393.5 kJ/mol)] - [(1*−74.87 kJ/mol)+(2*0 kJ/mol)] =<em> -802.27 kJ/mol</em>
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Explanation:

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Given the equation: 2H2 + O2 → 2H2O, how many moles of oxygen are needed to react with 13 moles of hydrogen
bulgar [2K]

Explanation:

2H2 + O2 = 2H2O

2mol. 1mol. 2mol

2mol reacts with 1mol

13mol reacts with x

x=<u>13mol</u><u> </u><u>×</u><u> </u><u>1mol</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2mol</u>

x= <u>13mol</u>

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>2mol

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6 0
2 years ago
Which of the following statements is correct about the force of gravity between two objects?
SVETLANKA909090 [29]

Answer: D

Explanation:

Shape and temperature have nothing to do with gravity. You could take a circle block and a square block and put it on the ground and they will stay there. Gravity is not weaker when objects are closer together because of Newtons law of gravitation.

3 0
3 years ago
A chemical factory is making soda ash (NA2CO3) from sodium bicarbonate. The production manager calculates they will make 80 tons
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Answer:

\boxed{\text{ 93 \%}}

Explanation:

\text{\% yield} = \dfrac{\text{actual yield}}{\text {theoretical yield}} \times\text{100 \%}

Data:

\begin{array}{rcr}\text{Actual yield} & = & \text{74.3 T}\\\text{Theoretical yield} & = & \text{80 T}\\\end{array}

Calculation:

\text{\% yield} = \dfrac{\text{74.3 T}}{\text {80 T}} \times\text{100 \%} = \textbf{93 \%}}\\\\\text{The percent yield was } \boxed{\textbf{93 \%}}

7 0
3 years ago
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