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Lostsunrise [7]
3 years ago
6

How many molecules of O2 are in 153.9 g O2?

Chemistry
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

molar mass of O2 molecule = 16*2=32g/mol

n= mole

To find the mole

n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

Nm= number of molecule

Nn = Number of mole

NA = number of Avogados

Nm= Nn * NA

Nm= 4.81 *6.02*10^23

Nm= 2.895*10^24

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
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The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

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