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3 years ago

Question 6

Chemistry

1 answer:

crimeas [40]3 years ago

8 0

Answer: There are $10.5\times 10^{23}$ formula units in 1.75 mol sample of NaCl.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of NaCl contains = $6.023\times 10^{23}$ formula units

Thus 1.75 mol of NaCl contains = $\frac{6.023\times 10^{23}}{1}\times 1.75=10.5\times 10^{23}$ formula units

Thus there are $10.5\times 10^{23}$ formula units in 1.75 mol sample of NaCl.

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Answer:

The temperature at which the given substance starts to boil is called boiling point.

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oth hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you opene

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Answer: Odor of ammonia would we detect first on the other side of the room.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

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3 years ago

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Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

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The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

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The value of x = 0.2077 M

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Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

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$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

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