⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️⚱️2 one.
Answer:
30.8 g of water are produced
Explanation:
First of all we need the equation for the production of water:
2H₂ + O₂ → 2H₂O
2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.
As we assume, the oxygen in excess, we determine the moles of H₂.
1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles
Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water
Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
5.5 L
Explanation:
First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:
- 10 g ÷ 44 g/mol = 0.23 mol
Then we <u>use the PV=nRT formula</u>, where:
- P = 1 atm & T = 293 K (This are normal conditions of T and P)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K
