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enyata [817]
1 year ago
11

How many molecules are in 3.5 moles of hydrogen

Chemistry
1 answer:
jek_recluse [69]1 year ago
7 0

Answer:3.5 moles * 6.02 X 10^23 particles/mole * 2 H atoms/particle of H2 = 42.14 X 10^^23 atoms

Explanation:

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the energy possessed by a body by its value of its position relative to others, stresses within itself, electric charge, and other factors.

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Select all statements that correctly describe the reactions of benzene. A. Benzene typically undergoes reactions in which the ar
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Answer:Benzene typically undergoes reactions in which the aromatic ring is preserved.B. Benzene typically reacts with electrophiles where an aromatic proton is substituted by the electrophile

Explanation:

The reactions of benzene are such that the aromatic ring is not destroyed. Addition reactions destroy the aromatic ring hence they aren't typical reactions of benzene. Benzene rings are attacked by electrophiles in which reaction a proton is substituted by the electrophile. Alkenes only undergo addition reaction and not electrophilic substitution reaction.

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3 years ago
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8 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the equation N2O4(g)
lora16 [44]

<u>Answer:</u> The value of equilibrium constant is 0.997

<u>Explanation:</u>

We are given:

Percent degree of dissociation = 39 %

Degree of dissociation, \alpha = 0.39

Concentration of N_2O_4, c = \frac{1mol}{1L}=1M

The given chemical equation follows:

                     N_2O_4\rightleftharpoons 2NO_2

<u>Initial:</u>                c             -

<u>At Eqllm:</u>         c-c\alpha      2c\alpha

So, equilibrium concentration of N_2O_4=c-c\alpha =[1-(1\times 0.39)]=0.61M

Equilibrium concentration of NO_2=2c\alpha =[2\times 1\times 0.39]=0.78M

The expression of K_{c} for above equation follows:

K_{c}=\frac{[NO_2]^2}{[N_2O_4]}

Putting values in above equation, we get:

K_{c}=\frac{(0.78)^2}{0.61}\\\\K_{c}=0.997

Hence, the value of equilibrium constant is 0.997

4 0
2 years ago
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