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2 years ago

5

A sample of helium gas occupies 2.65 L at 1.20 atm. What pressure would this sample of gas exert in a 1.50 L container at the sa

me temperature?

1 answer:

VladimirAG [237]2 years ago

5 0

Answer : The final pressure of the gas is, 2.12 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

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

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 1.20 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 2.65 L

$V_2$ = final volume of gas = 1.50 L

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

$1.20atm\times 2.65L=P_2\times 1.50L$

$P_2=2.12atm$

Therefore, the final pressure of the gas is, 2.12 atm

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Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp

sergiy2304 [10]

Explanation:

The given data is as follows.

$t_{m}$ = 30.0 sec, $t_{r1}$ = 5 min = $5 \times 60 sec$ = 300 sec

$t_{r2}$ = 12.0 min = $12 \times 60 sec$ = 720 sec

Formula for adjusted retention time is as follows.

$t'_{r} = t_{r} - t_{m}$

= 300 sec - 30.0 sec

= 270 sec

$t'_{r2}$ = 720 sec - 30 sec

= 690 sec

Formula for relative retention ( $\alpha$ ) is as follows.

$\alpha = \frac{t'_{r2}}{t'_{r1}}$

= $\frac{690 sec}{270 sec}$

= 2.56

Thus, we can conclude that the relative retention is 2.56.

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2 years ago

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made

natta225 [31]

Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L

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2 years ago

Gaseous ICl (0.20 mol) was added to a 2.0 L flask and allowed to decompose at a high temperature:

Ne4ueva [31]

Answer:

The Kc is 1.36 (but this is not an option, may be the options are wrong, or may be I was .. Thanks!)

Explanation:

Let's think all the situation.

2 ICl(g) ⇄ I₂(g) + Cl₂(g)

Initially 0.20 - -

Initially I have only 0.20 moles of reactant, and nothing of products. In the reaction, an x amount of compound has reacted.

React x x/2 x/2

Because the ratio is 2:1, in the reaction I have the half of moles.

So in equilibrium I will have

(0.20 - x) x/2 x/2

Notice that I have the concentration in equilibrium so:

0.20 - x = 0.060

x = 0.14

So in equilibrium I have formed 0.14/2 moles of I₂ and H₂ (0.07 moles)

Finally, we have to make, the expression for Kc and remember that must to be with concentration in M (mol/L).

As we have a volume of 2L, the values must be /2

Kc = ([I₂]/2 . [H₂]/2) / ([ICl]/2)²

Kc = (0.07/2 . 0.07/2) / (0.060/2)²

Kc = 1.225x10⁻³ / 9x10⁻⁴

Kc = 1.36

8 0

2 years ago

A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

2 years ago

When CO2(g) reacts with H2(g) to form C2H2(g) and H2O(g), 23.3 kJ of energy are absorbed for each mole of CO2(g) that reacts. Wr

kupik [55]

Answer:

$2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

Explanation:

As energy is absorbed therefore it is an endothermic reaction. Hence energy value should be written in the product side with a negative sign.

Reaction: $CO_{2}(g)+H_{2}(g)\rightarrow C_{2}H_{2}(g)+H_{2}O(g)$

C balance: $2CO_{2}(g)+H_{2}(g)\rightarrow C_{2}H_{2}(g)+H_{2}O(g)$

H and O balance: $2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)$

Here 2 moles of $CO_{2}$ react. So, energy absorbed during the reaction is $(2\times 23.3)$ kJ or 46.6 kJ

Energy balance: $2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

Balanced thermochemical equation:

$2CO_{2}(g)+5H_{2}(g)\rightarrow C_{2}H_{2}(g)+4H_{2}O(g)-46.6kJ$

3 0

2 years ago