As the question tells you, you need to use the formula
% mass= mass of solute/ mass of solution x 100
mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g
% mass= 30.0/ 300.0 x 100= 10%
answer is B
Answer: 9.68 x 10^10 grams.
Explanation:
Given that:
Mass of CO2 = ?
Number of molecules of CO2 = 2.2x10^9 molecules
Molar mass of CO2 = ? (let unknown value be Z)
For the molar mass of CO2: Atomic mass of Carbon = 12; Oxygen = 16
= 12 + (16 x 2)
= 12 + 32 = 44g/mol
Apply the formula:
Number of molecules = (Mass of CO2 in grams/Molar mass)
2.2x10^9 molecules = Z/44g/mol
Z = 2.2x10^9 molecules x 44g/mol
Z = 9.68 x 10^10g
Thus, the mass of 2.2x10^9 molecules of CO2 is 9.68 x 10^10 grams.
Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
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