Answer:
Search by reactants (P 2O 5, H 2O) and by products (H 3PO 4)
H2O + P2O5 → H3PO4
H2O + HNO3 + P2O5 → H3PO4 + N2O5
Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
<h3>
Reaction between oxygen and ethene</h3>
Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.
C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O
from the equation above;
3 moles of O₂ ---------> 2(18 g) of water
3.5 moles of O₂ ----------> x
![x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]](https://tex.z-dn.net/?f=x%20%3D%203.2%20%5Ctimes%20%5B%5Cfrac%7B2%20%5C%20moles%20%5C%20H_2O%7D%7B3%20%5C%20moles%20%5C%20O_2%7D%20%20%5D%20%5Ctimes%5B%20%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mole%20%5C%20H_2O%7D%20%5D)
Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
Learn more about reaction of ethene here: brainly.com/question/4282233
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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 formula of the chloride of M will be MCI2.