Answer:
The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio: The empirical formula is Fe2O3.
Answer:
B
101L
Explanation:
We use the ideal gas relation
PV = nRT
P = pressure = 101.3KPa
V = volume = ?
n = number of moles = 4.5moles
T = Temperature = 273.15K
R = molar gas constant = 8.314J/mol.k
Rearranging the equation to make V the subject of the formula yields :
V = nRT/P
= ( 4.5 × 8.314 × 273.15) ÷ 101.3
= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L
Answer:
% (COOK)2H2O = 37.826 %
Explanation:
mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O
∴ mass mix = 4.00 g
∴ mass (CaC2O4)H2O = 1.20 g
∴ Mw (COOK)2H2O = 184.24 g/mol
∴ Mw (CaC2O4)H2O = 146.12 g/mol
∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1
- % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100
⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)
⇒ mass (COOK)2H2O = 1.513 g
⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100
⇒ % (COOK)2H2O = 37.826 %
I’m not very good at this subject, however I have a idea of what it could be but if wrong My apologies “ Placed wires “.
Have a blessed day and God bless.