Answer:
Explanation:
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In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:
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Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:
b. .28 M KCI
Explanation:
Use the dilution formula.
C₁V₁ = C₂V₂
C₂ = C₁V₁/V₂
C₂ = 1.6*0.175 / 1.0
C₂ = 0.28
Answer:
A. Semi Permeable.
Explanation:
Cell membranes are semi permeable.
Answer:
63.6%
Explanation:
The given compound is:
N₂O;
The problem here is to find the percent composition of nitrogen in the compound.
First find the molar mass of the compound:
Molar mass of N₂O = 2(14) + 16 = 44g/mol
So;
Percentage composition of Nitrogen = 
x 100 = 63.6%
