<span>a. NaNO3: soluble
b. AgBr: insoluble
c. NH4OH: soluble
d. Ag2CO3: insoluble
e. NH4Br: soluble
f. BaSO4: insoluble
g. Pb(OH)2: soluble
h. PbCO3: insoluble</span>
1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??
22.4 x 0.001 / 1 => 0.0224 kg/mL
Answer:
M = 20.5 g/mol
Explanation:
Given data:
Volume of gas = 1.20 L
Mass of gas = 1.10 g
Temperature and pressure = standard
Solution:
First of all we will calculate the density.
Formula:
d = mass/ volume
d = 1.10 g/ 1.20 L
d = 0.92 g/L
Now we will calculate the molar mass.
d = PM/RT
0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K
M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm
M = 20.5 g/mol