I believe your answer is 2200000000000000. I hope this can help you :)
To remove magnesium oxide layer from the ribbon which may prevent or slow down the burning of magnesium ribbon.
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
Answer:
C. 0.4.
Explanation:
<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>
<em></em>
- no. of moles of acetic acid = 2, no. of moles of water = 3.
- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.
<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>
Answer:
lithium
Explanation:
this is because lithium has a valency of 1 and oxygen has a valency of 2 thereby exchanging valency to create Li²0