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LuckyWell [14K]
3 years ago
9

What is the empirical formula for a compound that contains 38.77% Cl and 61.23% O?

Chemistry
1 answer:
Leno4ka [110]3 years ago
3 0
W(Cl)=0.3877
w(O)=0.6123
M(Cl)=35.5 g/mol
M(O)=16.0 g/mol

ClₐOₓ
M(ClₐOₓ)=35.5a+16.0x

w(Cl)=35.5a/(35.5a+16.0x)
w(O)=16.0x/(35.5a+16.0x)

solve a system of two equations with two unknowns
35.5a/(35.5a+16.0x)=0.3877
16.0x/(35.5a+16.0x)=0.6123

a=2
x=7

Cl₂O₇ is the empirical formula
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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
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Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

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