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2 years ago

A solution composed of 30.6g NH3 in 81.3g of H20. Calculate the mole fraction for NH3 and H20

Chemistry

2 answers:

HACTEHA [7]2 years ago

8 0

Answer: NH3 = 9/5 H20= 9/2

Explanation:

Mr of NH3 = 17g/mol

Mr of H20 = 18g/mol

NH3 = 30.6g divide by 17g/mol = 1.8 mol

= 9/5

H20 = 81.3g divide by 18g/mol = 4.5 mol

= 9/2

irina [24]2 years ago

4 0

Answer:

0.286 moles

Explanation:

i hope this is helpful for you

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Answer:

Almost everything in our surroundings represent chemical change

Explanation:

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3 years ago

What is the change in atomic number when an atom emits a beta particle?

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3 years ago

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3 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.