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WINSTONCH [101]
3 years ago
5

In a 1.0-liter container there are, at equilibrium, 0.10 mole h2, 0.20 mole n2, and 0.40 mole nh3. what is the value of kc for t

his reaction at this temperature
Chemistry
1 answer:
bogdanovich [222]3 years ago
6 0
The reaction involved here would be written as:

2N2 + 3H2 = 2NH3

The equilibrium constant of a reaction is the ratio of the concentrations of the products and the reactants when in equilibrium. The expression for the equilibrium constant of this reaction would be as follows:

Kc = [NH3]^2 / [N2]^2[H2]^3
Kc = 0.40^2 / (0.20)^2 (0.10)^3 
Kc = 4000
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Answer:

Moon rocks contain few volatile substances (e.g. water), which implies extra baking of the lunar surface relative to that of Earth. The relative abundance of oxygen isotopes on Earth and on the Moon are identical, which suggests that the Earth and Moon formed at the same distance from the Sun.

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How are cellular respiration and photosynthesis related?
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Enter the molecular equation representing aqueous nitric acid and aqueous ammonia reacting. express your answer as a balanced mo
PtichkaEL [24]
Aqueous nitric acid and aqueous ammonia reacts to form ammonium nitrate 
 HNO₃(aq)+NH₃(aq) = NH₄NO₃(aq)
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 H⁺(aq) + NH₃ = NH₄⁺ (aq)

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3 years ago
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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0
3 years ago
A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.
lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

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3 years ago
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