All categories

2 years ago

A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp

Chemistry

2 answers:

miss Akunina [59]2 years ago

5 0

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

Fynjy0 [20]2 years ago

4 0

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

You might be interested in

What is the equation for figuring out the Atomic Mass

zhenek [66]

Answer:

Atomic mass = mass of protons + mass of neutrons

Explanation:

In an element atomic mass of an atom can be calculated by adding the mass of protons and neutrons

I hope this helps! :)

8 0

2 years ago

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at

kozerog [31]

Answer:

20 g/mol

Explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r = \frac{1 }{\sqrt{M}}$

If you have two gases, the ratio of their rates of diffusion is

$\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}$

Squaring both sides, we get

$(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}$

Solve for M₂:

$M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}$

$M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}$

$= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}$

7 0

3 years ago

How much heat is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling? Specif

NNADVOKAT [17]

Answer:

c. 157 KJ

Explanation:

Q= mC dT dT= 100°C(boiling point) - 25°C=75°C

Q= (500 g * 4.184 J/g °C * 75 °C)

Q= 156900 J= 157 KJ

6 0

2 years ago

Write a balanced chemical equation for the neutralization reaction between phosphoric acid and lithium hydroxide.

Akimi4 [234]

Hope that helps!!!!!!!

5 0

2 years ago

10.

Hoochie [10]

Answer:

Salt and water

Explanation:

The reaction of an acid and a base forming a salt and water.

3 0

2 years ago