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2 years ago

Which diagram shows the correct way to represent an ionic compound of magnesium oxide?

Chemistry

1 answer:

Daniel [21]2 years ago

8 0

The answer is ................................ c

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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

2 years ago

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!!15 points!!Which of the following phrases describes how the position of an electron

ki77a [65]

Answer:

A. The farther an electron is from the nucleus, the greater its energy.

Explanation:

8 0

2 years ago

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How many ATP are produced from a fatty acid that is 14 carbons long?

fenix001 [56]

Answer:

92 ATP

Explanation:

Fatty acid oxidation results in the formation of large number of ATP molecules. Three important process of fatty acid are activation of the fatty acid, beta oxidation and entry of acetyl CoA in Krebs cycle.

14 carbon fatty acid is Miristic acid. The complete oxidation of Miristic acid results in the formation of 7 acetyl CoA + 6NADH and $6FADH_2$

1 Acetyl CoA gives 10 molecules of ATP then 7 acetyl CoA gives 70 molecules of ATP.

1NADH = 2.5 ATP, 6NADH = 15 ATP.

$1FADH_2$ = 1.5 ATP, $6FADH_2$ = 9 ATP.

2 ATP has been consumed in the activation of fatty acid.

Total ATP = 70+15+9-2

=92 ATP.

Thus, the total ATP generated from the oxidation of 14 carbon fatty acid is 92.

3 0

2 years ago

The question is on the pic, thanks :)

Inessa05 [86]

It’s the BOA not the dog or kangaroo

8 0

1 year ago

How many hydrogen atoms are in the following molecule of ammonium sulfide? (NH4)2S

Lelechka [254]

The are eight hydrogen atoms in ammonium sulfide because there are 2 molecules of ammonium.

3 0

2 years ago

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