All categories

3 years ago

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure has its volume increased

Chemistry

1 answer:

allochka39001 [22]3 years ago

5 0

Answer:

2.5 atm

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15k

P2 = 2.0atm

P1 = ?

From general gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P2 × V2 × T1 = P1 × V1 × T2

P1 = (P2 × V2 × T1) / (V1 × T2)

P1 = (2.0 × 34 × 318.15) / (28 × 308.15)

P1 = 21634.5 / 8628.2

P1 = 2.5 atm

The initial pressure of the gas is 2.5atm

You might be interested in

Two students made the Lewis dot diagrams of H2O. The diagrams are as shown.

netineya [11]

Answer:

Only student A

Explanation:

don’t worry I got a 100

7 0

4 years ago

Read 2 more answers

What is the relationship between chemical equilibrium and the rates of forward and reverse reaction?

Nataly_w [17]

Answer:

Explanation:

the rate of a forward process must be exactly balanced by the rate of the reverse process.

3 0

3 years ago

Read 2 more answers

A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, it’s water level rises to 20.3 ml

MrRa [10]

Answer:

20.3-17.5=2.8ml 1ml=1cm3 vol= 2.8cm3

Explanation:

4 0

3 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

PLEASE HELP ASAP!!

DanielleElmas [232]

lights bullet electric heater radio Google

8 0

3 years ago

Read 2 more answers