What pressure is required to reduce 63 mL of a gas at standard conditions to 20 mL at the temperature of 24◦C? Answer in units of atm, please.

3.43atm

Explanation:

Using the combined gas law equation as follows:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information in this question,

P1 = 1 atm (standard pressure)

P2 = ?

V1 = 63mL

V2 = 20mL

T1 = 273 K (standard temperature)

T2 = 24°C = 273 + 24 = 297K

Using P1V1/T1 = P2V2/T2

1 × 63/273 = P2 × 20/297

63/273 = 20P2/297

Cross multiply

297 × 63 = 273 × 20P2

18711 = 5460P2

P2 = 18711/5460

P2 = 3.426

P2 = 3.43atm

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles