<span>a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+
b. Should form 5.11 X 10^-3 mol Cu(OH)2
c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc
d. 5.11 X 10^-3 mol Mg X 24.3 g/mol = 0.124 grams Mg</span>
Answer:
520 kg
Explanation:
Let's consider the combustion of isooctane.
C₈H₁₈(l) + 12.5 O₂(g) → 8 CO₂(g) + 9 H₂O(l)
We can establish the following relations:
- 1 mL of C₈H₁₈ has a mass of 0.690 g (ρ = 0.690 g/mL).
- The molar mass of C₈H₁₈ is 114.22 g/mol.
- The molar ratio of C₈H₁₈ to O₂ is 1:12.5.
- The mole fraction of O₂ in air is 0.21.
- The molar mass of air is 28.96 g/mol.
50 L of isooctane require the following mass of air.

Answer:
0.1357 M
Explanation:
(a) The balanced reaction is shown below as:

(b) Moles of
can be calculated as:
Or,
Given :
For
:
Molarity = 0.1450 M
Volume = 10.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 10×10⁻³ L
Thus, moles of
:
Moles of
= 0.00145 moles
From the reaction,
1 mole of
react with 2 moles of NaOH
0.00145 mole of
react with 2*0.00145 mole of NaOH
Moles of NaOH = 0.0029 moles
Volume = 21.37 mL = 21.37×10⁻³ L
Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M