C. The number of moles of H in 0.109 mole of N₂H₄ is 0.436 mole
D. The number of moles of H in 34 moles of C₁₀H₂₂ is 748 moles
<h3>C. How to determine the number of mole of H in 0.109 mole of N₂H₄</h3>
1 mole of N₂H₄ contains 4 moles of H
Therefore,
0.109 mole of N₂H₄ will contain = 0.109 × 4 = 0.436 mole of H
<h3>D. How to determine the number of mole of H in 34 mole of C₁₀H₂₂</h3>
1 mole of C₁₀H₂₂ contains 22 moles of H
Therefore,
34 mole of C₁₀H₂₂ will contain = 34 × 22 = 748 mole of H
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Answer:
Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule
Explanation:
An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.
However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.
The attacking nucleophile in this case may be water or CH3OH
Answer: Mass of
produced in this reaction was 6.56 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass or reactants = Mass of
+ mass of
= 16.00 + 64.80 = 80.80 g
Mass of products = mass of aqueous solution + mass of
+ = 74.24 + x g
Mass or reactants = Mass of products
80.80 g = 74.24 + x g
x = 6.56 g
Thus mass of
produced in this reaction was 6.56 grams