Answer:
See Explanation
Explanation:
For SF6;
Since;
1.25 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55
For SF4;
Since;
1.88 g of S corresponds to 4.44g of F
1 g of sulphur corresponds to 1 * 4.44/ 1.88 = 2.36
Hence;
Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4
=
3.55/2.36 = 1.5
Hence the law of multiple proportion is obeyed here.
Answer:
roar
Explanation:
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Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L