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Schach [20]
3 years ago
14

When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I

f the heat capacity of the bomb calorimeter is 1.229 kJ/°C, what is the heat of combustion for the unknown hydrocarbon, in kJ/mol?
Chemistry
1 answer:
Andreyy893 years ago
8 0

Answer:

S A T. NV

Explanation:

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The three end stages of stars are Red Giant Phase, Second Red Giant Phase and White Dwarf Phase correct?
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3 years ago
According to the National Institute of Health, ventricular tachycardia is a fast, regular beating of the ventricles that may las
jarptica [38.1K]

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1 makes the most sense

Explanation:

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3 years ago
Repeating many forms a crystal structure inside each of these repeated units is the ?
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7 0
3 years ago
Read 2 more answers
If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?
ICE Princess25 [194]

<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}

where,

k = rate constant = ?

t = time taken = 1.52 hrs

[A_o] = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}

To calculate the half life period of first order reaction, we use the equation:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life period of first order reaction = ?

k = rate constant = 0.2098hr^{-1}

Putting values in above equation, we get:

t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs

Hence, the half life of the sample of silver-112 is 3.303 hours.

6 0
3 years ago
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