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3 years ago

Chemistry: Gas Laws Problems!

Chemistry

1 answer:

Ronch [10]3 years ago

7 0

Answer : The final volume at STP is, 1000 L

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 1520 mmHg = 2 atm (1 atm = 760 mmHg)

$P_2$ = final pressure at STP = 1 atm

$V_1$ = initial volume = 500.0 L

$V_2$ = final volume at STP = ?

Now put all the given values in the above formula, we get:

$2atm\times 500.0L=1atm\times V_2$

$V_2=1000L$

Therefore, the final volume at STP is, 1000 L

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Answer:

I'm unaware of what but maybe hot sauce?

Explanation:

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3 years ago

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How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102

Ede4ka [16]

Answer: $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

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.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles$

1 mole of hydrogen $(H_2)$ = $2\times 6.023\times 10^{23}=12.05\times 10^{23}$ atoms

17.5 mole of hydrogen $(H_2)$ = $\frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25}$ atoms

There are $2.12\times 10^{25}$ atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0

3 years ago

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$