Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of H = 18 %
Percentage of N = 82 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 18 g
Mass of N = 82 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.
For Hydrogen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of H : N = 3 : 1
Hence, the empirical formula for the given compound is
Nonpoint source pollution
Answer:
Assuming that the atmospheric pressure is standard, the temperature will rise steadily until it reaches 100°C
1. Answer:
1.0 × 10–9 M OH–
Explanation:
pH = -Log[H+]
pOH = -Log[OH-]
But;
pH + pOH = 14
Therefore;
[H+] + [OH-] = 1.0 × 10^-14 M
Therefore;
[OH-] = 1.0 × 10^-14 M - (1.0 × 10^–5 M)
= 1.0 × 10^-9 M OH–
2. Answer;
pH = 7.28
Explanation;
pH = -Log[H3O+]
Given;
[H3O+] = 5.2 × 10^–8 M
Therefore;
pH = - log [5.2 × 10^–8 M]
= 7.28
The pH is 7.28
Explanation:
Mass of compounds = Moles of compound × Molecular mass of compound
a) Moles of LiCl = 2.345 mol
Molecular mass of LiCl = 42.5 g/mol
Mass of 2.345 moles of LiCl = 2.345 mol × 42.5 g/mol = 99.6625 g
b) Moles of acetylene = 0.0872 mol
Molecular mass of acetylene= 26 g/mol
Mass of 0.0872 moles acetylene= 0.0872 mol × 26 g/mol = 2.2672 g
c) Moles of sodium carbonate= 
Molecular mass of sodium carbonate= 106 g/mol
Mass of sodium carbonate
= 
= 3.498 g
d) Moles of fructose = 
Molecular mass fructose= 180 g/mol
Mass of fructose
= 
e) Moles of 
Molecular mass of 
Mass of fructose
= 
