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maks197457 [2]
1 year ago
15

15 POINTS!! does anyone know the answer to this? is it A?

Chemistry
2 answers:
Norma-Jean [14]1 year ago
7 0

Answer:

b.2

Explanation:

lakkis [162]1 year ago
5 0

Answer:

Mole fraction of water(H₂O) is 0.5 and the mole fraction of CH₃OH(Methanol) is 0.5.

Explanation:

Greetings !

The molecular weight of CH₃OH(Methanol)=32g/mol

The number of moles of CH₃OH=

\frac{128g}{32g/mol}

=4moles

The molecular weight of water H₂O =18g/mol

The number of moles of=

\frac{72g}{18g/mol}

=4moles

Total number of moles in the solution =4mol + 4mol

=8mol

Mole fraction Methanol CH₃OH=

\frac{4mol}{8mol}

=0.5mol

Hope it helps!

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\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

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From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

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The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

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\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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