Answer:
see calculations in explanation
Explanation:
percent = part/total x 100%
part = ∑ atomic mass of element
- hydrogen = 1.008 amu (atomic mass units)
- carbon = 12.011 amu
- nitrogen = 14.007 amu
total = ∑ molecular mass of compound
= H amu + C amu + Namu
= 1.008 amu + 12.011 amu + 14.007 amu
= 27.026 amu
%H = (1.008amu/27.026amu)100% = 3.730%
%C = (12.011amu/27.026amu)100% = 44.442%
%N = (14.007amu/27.026amu)100% = 51.827%
Check results ∑%values = 100%
3.730% + 44.442% + 51.827% = 99.999% ≅ 100%
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Answer:
0.88g
Explanation:
The reaction equation:
2NaI + Cl₂ → 2NaCl + I₂
Given parameters:
Mass of Sodium iodide = 2.29g
Unknown:
Mass of NaCl = ?
Solution:
To solve this problem, we work from the known to the unknown.
First find the number of NaI from the mass given;
Number of moles = 
Molar mass of NaI = 23 + 126.9 = 149.9g/mol
Now insert the parameters and solve;
Number of moles = 
= 0.015mol
So;
From the balanced reaction equation;
2 moles of NaI produced 2 moles of NaCl
0.015mole of NaI will produce 0.015mole of NaCl
Therefore;
Mass = number of moles x molar mass
Molar mass of NaCl = 23 + 35.5 = 58.5g/mol
Now;
Mass of NaCl = 0.015 x 58.5 = 0.88g
Answer:
20.3 % NaCl
Explanation:
Given data:
Mass of solute = 45.09 g
Mass of solvent = 174.9 g
Mass percent of solution = ?
Solution:
Mass of solution = 45.09 g + 174.9 g
Mass of solution = 220 g
The solute in 220 g is 45.09 g
220 g = 2.22 × 45.09
In 100 g solution amount of solute:
45.09 g/2.22 = 20.3 g
Thus m/m% = 20.3 % NaCl
