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2 years ago

a gas exerts a pressure of 1.00 atm at 273 K. At what tempreture the gas exerts a pressure of 4.00 atm.

Chemistry

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

Considering Gay-Lussac's law, a gas exerts a pressure of 4.00 atm at 1,092 K.

<h3>Gay-Lussac's law</h3>

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, gas pressure decreases.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T}=k$

where P= pressure, T= temperature, k= Constant

This law states that the ratio of pressure to temperature is constant.

Studying an initial state 1 and a final state 2, it is fulfilled:

$\frac{P1}{T1}=\frac{P2}{T2}$

<h3>Final temperature</h3>

In this case, you know:

P1= 1 atm
T1= 273 K
P2= 4 atm
T2= ?

Replacing in Gay-Lussac's law:

$\frac{1 atm}{273 K}=\frac{4 atm}{T2}$

Solving:

$T2\frac{1 atm}{273 K}=4 atm$

$T2=\frac{4 atm}{\frac{1 atm}{273 K}}$

Finally, a gas exerts a pressure of 4.00 atm at 1,092 K.

Learn more about Gay-Lussac's law:

brainly.com/question/4147359

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How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .