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USPshnik [31]
2 years ago
7

Oxyacetylene torches produce such high temperature that they are often used to weld and cut metal. When 1.53 g of acetylene (C2H

2) is burned in a bomb calorimeter with a heat capacity of 10.69 kJ/K, the temperature increases from 20.486°C to 44.688°C. What is ΔE (in kJ/mol) for this combustion reaction? Enter to 0 decimal places.
Chemistry
1 answer:
hodyreva [135]2 years ago
7 0

Answer: 4403kJ/mole

Explanation:

First we have to calculate the heat absorbed by bomb calorimeter  

Formula used :

q_b=c_b\times (T_{final}-T_{initial})

q_b = heat absorbed by calorimeter = ?

c_b = specific heat of = 10.69 kJ/K

T_{final} = final temperature = 44.688^oC=(273+44.688)K=317.688K

T_{initial} = initial temperature = 20.486^oC=(273+20.486)K=293.486K

q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.

Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ

So, 26.04 g/mole of acetylene releases heat of combustion \frac{258.7}{1.53}\times 26.04=4403kJ/mole

Therefore, the heat of combustion of acetylene is, 4403kJ/mole

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Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

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6 0
2 years ago
The percent composition by mass of a compound is 76.0% c, 12.8% h, and 11.2% o. the molar mass of this compound is 284.5 g/mol.
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You have a few steps to solve this one. First, we'll find the molar mass by percentage of each element in the molecule. Then, we'll divide each of those relative masses by the atomic mass of each element. The number of times the mass divides into the relative mass is the number of atoms of that element in the molecule:

C: 284.5 x .76 = 216.22
H: 284.5 x .128= 36.416
O: 284.5 x .112 = 31.864.

Now we divide out each element's atomic mass (from the periodic table). it's okay if they're approximated from the decimal answer.
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H: 36.416 ÷ 1.008 ≈36
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The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O




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