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GuDViN [60]
2 years ago
15

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the

water rose by 4.20 Ce … lsius. If the heat capacity of the bomb plus water was 10.4 kJ/degree Celsius, calculate the molar heat of combustion of methanol
Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
4 0

The formula for calculating the amount of energy or heat released is:

ΔH = C ΔT

where ΔH is heat of combustion, C is heat capacity, while ΔT is change in temperature

ΔH = 8.69 kJ / °C * (5.14°C)

ΔH = 44.67 kJ

Then we calculate the moles of CH3OH which has molar mass of 32.04 g/mol:

moles = 1.922 / 32.04 = 0.05999 mol

SO the molar heat of combustion is:

ΔHm = 44.67 kJ / 0.0599875 mol

ΔHm = 744.60 kJ / mol

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Naddika [18.5K]

Answer:

Percentage of carbon:

{ \tt{ =  \frac{24}{30}  \times 100\%}} \\  = 80\%

Percentage of hydrogen:

{ \tt{ =  \frac{6}{30}  \times 100\%} } \\  = 20\%

8 0
3 years ago
Equation for questions 1-4:
Karolina [17]

Answer:

5.625 moles of oxygen, O₂.

Explanation:

The balanced equation for the reaction is given below:

4Al + 3O₂ —> 2Al₂O₃

From the balanced equation above,

4 moles of Al reacted with 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O₂.

Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.

Thus, 5.625 moles of O₂ is needed for the reaction.

3 0
3 years ago
Each step in the following process has a yield of 70.0%. CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl The CCl4 formed in the first st
Nadusha1986 [10]

Answer:

n(HCl)=1.96 mol

Explanation:

CH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HCl

With ideal yields we will end up with 4 moles of HCl.

With 70% yields on every stage

n(HCl)=0.7*0.7*4=1.96 mol

8 0
3 years ago
Explain the Doppler effect using light. Why is the Doppler effect important in astronomy?
Alexeev081 [22]
Doppler effect is the compression or extension of a sound wave, which causes a change in its wavelength / frequency (and so its sound).
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8 0
3 years ago
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How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are
Otrada [13]

Answer:

Option 3. The catalyst does not affect the enthalpy change (\Delta H_\text{rxn}) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction (\Delta H_\text{rxn}) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change \Delta H_\text{rxn} of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction \Delta H_\text{rxn} measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

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