I cannot see your question to help you... sorry
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
so basically
some fuels have an impurity in them which is sulfur.
When the fuel undergoes combustion, the sulfur reacts with oxygen in the air to form sulfur dioxide.
the sulfur dioxide reacts with water vapour in the air to form sulfurous acid, which is a type of acid rain.
Also
the high pressures inside a car engine may cause nitrogen and oxygen in the air to react and form oxides of nitrogen. the most common compounds formed inside car engines are NO (nitrogen oxide) and NO2 (nitrogen dioxide)
Answer:
You will need 12 moles of F2 if you want to make 8 moles of AlF3.
Explanation:
It takes 3 moles F2 to make 2 moles of AlF3 (this will be our mole ratio)
2 moles AlF3/3 moles F2 =8 moles AlF3/x moles AlF3.
x=12 moles AlF3
