Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
The percentage composition of this compound : 40%Ca, 12%C and 48%O
<h3>Further explanation</h3>
Given
20.0 g of calcium,
6.0 g of carbon,
and 24.0 g of oxygen.
Required
The percentage composition
Solution
Total mass of compound :
=mass calcium + mass carbon + mass oxygen
=20 g + 6 g + 24 g
=50 g
Percentage composition :
You have to use the equation F=ma and solve for m to get m=F/a.
m=mass in kg
F=force (in this case 350N)
a=acceleration (in this case 10m/s²)
when you plug everything in you should find that m=35kg
I hope this helps.
Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar