As,
CuCO₃ ⇆ Cu²⁺ + CO₃²⁻
So,
Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]
Or,
Ksp = [Cu²⁺] [CO₃²⁻]
As,
Ksp = 1.4 × 10⁻¹⁰
So,
1.4 × 10⁻¹⁰ = [x] [x]
Or,
x² = 1.4 × 10⁻¹⁰
Or,
x = 1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
x = 1.18 × 10⁻⁵ mol/L × 123.526 g/mol
x = 1.45 × 10⁻³ g/L
The answer is the letter C
Answer:
1. three-dimensional bonding
Diamond
2. Hardest natural substance
diamond
3. Used as lubricant
Graphite
4. nonconductor
Diamond
5. Weak, planar bonds
Graphite
6.Carbon black or soot
Amorphous
Explanation;
sorry im late, hopefully this can help somebody :)
The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
If you like my answer, please vote me a 'brainliest' - trying to improve my rank :-)
