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adell [148]
2 years ago
10

How many moles of ethane gas, C2H6(g), will occupy 15.0 L at 30 degrees Celsius and 5.0 atm?

Chemistry
1 answer:
Alenkinab [10]2 years ago
8 0

Answer:

3 mol

Explanation:

Given data:

Volume of ethane = 15.0 L

Temperature = 30°C

Pressure = 5.0 atm

Number of moles of ethane = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

30+273 = 303 K

5.0 atm × 15.0 L = n×0.0821 atm.L/ mol.K   × 303 K

75 atm.L =  n× 24.87 atm.L/ mol

n = 75 atm.L / 24.87 atm.L/ mol

n = 3 mol

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7. A nonpolar covalent bond (i.e., pure covalent) would form in which one of the following pairs of atoms? A) Na Cl B) H Cl C) L
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E) Br Br

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Covalent bond -

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8 0
2 years ago
Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di
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Answer : The correct rate law for the reaction is,

\text{Rate}=k[CHCl_3][Cl_2]^{1/2}

Explanation :

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For the given chemical equation:

CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)

Rate law expression for the reaction:

\text{Rate}=k[CHCl_3]^a[Cl_2]^b

where,

a = order with respect to CHCl_3

b = order with respect to Cl_2

Expression for rate law for first observation:

0.0035=k(0.010)^a(0.010)^b ....(1)

Expression for rate law for second observation:

0.0069=k(0.020)^a(0.010)^b ....(2)

Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

0.027=k(0.040)^a(0.040)^b ....(4)

Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

Dividing 2 from 3, we get:

\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}

Calculation used :

1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}

Thus, the rate law becomes:

\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}

7 0
3 years ago
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