The moles of Ba(OH)2 that is required to react with 117 HBr is calculated as below
find the moles of HBr used
mass/ molar mass = 117 g/ 80.9 g/mol = 1.446 moles
write the reacting equation
Ba(OH)2 + 2 HBr = BaBr2 + 2 H2O
by use of mole ratio of Ba(OH)2 : HBr which is 1:2 the moles of Ba(OH)2 is therefore
= 1.446 moles x1/2 = 0.723 moles of Ba(OH)2
Both compare the rate of population growth.
Answer:
839 g
Explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 81.97 209.94
NaAlO₂ + … ⟶ Na₃AlF₆ + …
Mass/g: 2150
1. Calculate the <em>moles of Na₃AlF₆
</em>
Moles of Na₃AlF₆ = 2150 × 1/209.94 Do the operation
Moles of Na₃AlF₆ = 10.24 mol Na₃AlF₆
2. Calculate the <em>moles of NaAlO</em>₂
The molar ratio is 1 mol NaAlO₂:1 mol Na₃AlF₆
Moles of NaAlO₂ = 10.24 × 1/1 = 10.24 mol NaAlO₂
3. Calculate the <em>mass of NaAlO₂
</em>
Mass of NaAlO₂ = 10.24 × 81.97 Do the multiplication
Mass of NaAlO₂ = 839 g
The answer is A. Voltaic cells couple oxidation and reduction half reactions to produce electromotive force from chemical potential energy. Electrolytic cells do the opposite in that they require electromotive force to store chemical energy.
