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3 years ago

Change in enthalpy when 11.2 dm3 of helium at ntp is heated in a cylinder to 100 degrees celsius

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Enthalpy change = 333.3 J

Explanation:

Given data:

Volume of helium = 11.2 dm³ or 11.2 L

Temperature = 100 °C (100+273= 373 K)

Enthalpy change = ?

Solution:

Normal temperature of room = 20°C (20+273 = 293 K)

one mole of gas occupy volume = 22.4 L

ΔT = final temperature - initial temperature

ΔT = 373K - 293K

ΔT = 80K

Number of moles of gas = 11.2 L/22.4 l

Number of moles of gas = 0.5 mol

Enthalpy change = nRΔT

Enthalpy change = 0.5 mol × 8.3j/mol.k ×80 K

Enthalpy change = 333.3 J

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Answer:

1.16 mL

Explanation:

Assume the data show that 25 pennies have a total volume of 29.00 mL.

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3 years ago

Consider the mechanism. step 1:step 2:overall:2a⟶bb c⟶d2a c⟶dslowfast determine the rate law for the overall reaction, where the

Ne4ueva [31]

Answer:

Rate: R = k [A]²

Explanation:

The rate law equation for a given complex chemical reaction is given as the product of molar concentrations of the reactant raised to the power their respective partial reaction orders. The sum of the partial reaction orders is equal to the overall order of reaction.

For a complex reaction, the rate law is generally determined by the slowest step, which is known as the rate-determining step.

<u>Given reaction mechanism:</u>

Step 1: 2 A⟶ B, slow step

Step 2: B+ C⟶ D, fast step

Overall: 2 A + C ⟶ D

In this given reaction mechanism, the <em>step 1 is the slow step and thus the rate determining step.</em>

Therefore, the rate law for the given mechanism is:

Rate: R = k [A]²

Here, k is the overall rate constant

Also, the overall order of the reaction = 2

Therefore, the given chemical reaction is said to be second order reaction.

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3 years ago

When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I

Alexus [3.1K]

Answer:

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

$Q=C\times \Delta T$

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Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

$\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol$

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