Question 4 2 pts
669.0 mL of oxygen are collected over water at 17.0 °C and a total
pressure of 785.0 mm of mercury. What is the volume (in mL) of dry
oxygen at 60.0 °C and 847.0 mmHg pressure?
Question 5
2 pts

711.96 mL

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (mL)

V2 = final volume (mL)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 785.0 mmHg

P2 = 847.0 mmHg

V1 = 669.0 mL

V2 = ?

T1 = 17.0 °C = 17 + 273 = 290K

T2 = 60.0 °C = 60 + 273 = 333K

Using P1V1/T1 = P2V2/T2

785 × 669/290 = 847 × V2/333

525165/290 = 847 V2/333

1810.91 = 2.54 V2

V2 = 1810.91 ÷ 2.54

V2 = 711.96 mL

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer