Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.
The reaction produces 2.93 g H₂.
M_r: 133.34 2.016
2Al + 6HCl → 2AlCl₃ + 3H₂
<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃
<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂
<em>Mass of H₂</em> = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6
