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Margarita [4]
2 years ago
14

How much heat is released when 15.7g of methane (c2h6) is combusted if the enthalpy of the reaction is - 1560.7 kj

Chemistry
1 answer:
lyudmila [28]2 years ago
5 0

- 407.4 kJ of heat is released.

<u>Explanation:</u>

We have to write the balanced equation as,

2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O

Here 2 moles of ethane reacts in this reaction.

Now we have to find out the amount of ethane reacted using its given mass and molar mass as,

2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆

Heat released = ΔH × given mass / 60.14

                        = - 1560. 7 kj ×15.7 g / 60. 14 g  = -407. 4 kJ

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Answer:

Explanation:

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The specific heat of liquid water is 4.184 j/g· ?c. calculate the energy required to heat 10.0 g of water from 26.5?c to 83.7?c.
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Energy required=mass*specific heat*temperature change
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5 0
3 years ago
Answer the lab question (“What is the effect of temperature on the solubility of a solid in a liquid?”) with a hypothesis:
valkas [14]

Explanation:

When experimenting, the best hypothesis to develop would be a null hypothesis (H₀). A null hypothesis is a statement indicating no change or effect.  In this case, it would be;

“There is no effect of temperature on the solubility of a solid in a liquid”

An alternative hypothesis (Hₐ) would be;

“There is an effect of temperature on the solubility of a solid in a liquid”

In this experiment, the null hypothesis would be rejected and the alternative would be accepted. This is because the experiment would show that increased temperatures of the liquid increases solubility of the solid in the liquid.

8 0
3 years ago
Read 2 more answers
Have you ever thought about how an MP3 player works? What form of energy is used to power an MP3 player? What form of energy do
Irina-Kira [14]

Answer:

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4 0
3 years ago
g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con
Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

t_{1/2}=\frac{0.693}{k}} .....(1)

Rate law expression for first order kinetics is given by the equation:

t=\frac{2.303}{k}\log\frac{[A_o]}{[A]} ......(2)

where,  

k = rate constant

t_{1/2} =Half life of the reaction = 2.42\times 10^3 s

t = time taken for decay process = ?

[A_o] = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of [A_o]

[A]=\frac{66.8}{100}\times 0.163 M=0.108884 M

k=\frac{0.693}{2.42\times 10^3 s}

t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}

t = 1,409.19 s

1 minute = 60 sec

t=\frac{1,409.19 }{60} min=23.49 min

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

6 0
2 years ago
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