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Elodia [21]
3 years ago
10

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 x 102. What is the form of the equilibri

um constant?
. [NH3]2

[N2][H2]3


. [NH3]_

[N2][H2]3


[NH3]2

[N2][H2]


[NH3]2

[N2][3H2]3
Chemistry
1 answer:
madam [21]3 years ago
5 0

Answer:

First option:

[NH3]2  / [N2][H2]3

Explanation:

First of all we need to determine the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

1 mol of nitrogen can combine with 3 moles of hydrogen to produce 2 moles of ammonia.

Remember that Kc has to involve Molar concentrations of products / reactants

Concentration must be elevated to the stoichiometry coefficient

Kc for the reaction is:  [NH₃]²  / [N₂] . [H₂]³

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Balance this equation: Al+ HNO3 ---- H2 + Al(NO3)3​
romanna [79]

Answer:

2Al+ 6HNO3 ---- 3H2 + 2Al(NO3)3​

Explanation:

Put coefficient a,b,c, and d for calculation:

a Al + b HNO3 = c H2 + d Al(NO3)3

for Al: a = d

for H: b = 2c

for N: b = 3d

for O: 3b = 9d

Suppose a=1, then d=1, b=3, c=3/2

multiply 2 to make all natural number, a=2, then b=6, c=3, d=2

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Answer:

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2 years ago
Whuch expression is equal to the number of grams in 2.43 kg
schepotkina [342]
2.43*1000. 

There are 1000 grams in a kilogram
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3 years ago
Suppose an object is moving through space in a straight line. What could cause the object to start moving in a circle?
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8 0
3 years ago
Read 2 more answers
A substance is analyzed and found to contain 85.7% carbon and 14.3% hydrogen by weight. A gaseous sample of the substance is fou
atroni [7]

Answer:

The empirical formula of the compound is CH2

Explanation:

<u>Step 1:</u> Data given

A substance contains 85.7 % carbon and 14.3 % hydrogen.

The substance has a density of 1.87 g/L

1 mol occupies 22.4 L

Molar mass of carbon = 12 g/mol

Molar mass of hydrogen = 1.01 g/mol

<u>Step 2</u>: Calculate molar mass of the substance

Since 1 mol occupies 22.4 L;

1 mol of this substance = 1.87g/L *22.4 = 41.888 grams

This means the molar mass of the substance is 41.888 g/mol

<u>Step 3:</u> Calculate mass of carbon:

85.8 % is carbon

this means 41.888 * 0.858 = 35.94 grams

<u>Step 4: </u>Calculate moles of carbon

moles C = mass C/ Molar mass C

Moles C = 35.94 grams / 12 g/mol

Moles C = 2.995 moles

<u>Step 5:</u> Calculate mass of hydrogen:

14.3 % is hydrogen

this means 41.888 * 0.143 = 5.99 grams

<u>Step 6 :</u>Calculate moles of hydrogen

Moles H  = 5.99 grams / 1.01 g/mol

Moles H = 5.93 moles

<u>Step 7: </u>Calculate  mol ratio

Ratio C:H = 1:2

The empirical formule = CH2

<u>Step 8</u>: calculate molar formule

Molar mass of empirical formule = 14.02 g/mol

n = Molar mass of substance / molar mass of empirical formule

n = 41.888 / 14.02 = 3

This means we have to multiply the empirical formula by 3

3*(CH2) = C3H6

C3H6 can be propene or cyclopropane

5 0
3 years ago
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