Answer:
The rate of a chemical reaction is directly proportional to the concentration. The rate of a chemical reaction increases as the substrate concentration increases and thus, concentration of the substrate in an enzyme-controlled chemical reaction increases with time.
Explanation:
The rate of a chemical reaction is directly proportional to the concentration.
The reaction rate increases with increasing substrate concentration, but levels off at a much lower rate. By increasing the enzyme concentration, the maximum reaction rate greatly increases.
Generally, the rate of a chemical reaction increases as the substrate concentration increases, and thus, concentration of the substrate in an enzyme-controlled chemical reaction increases with time.
Answer:
Ethanol = 102.29 g
Carbon Dioxide = 97.71 g
Solution:
The Balance Chemical Equation for the fermentation of Glucose is as follow,
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂
Mass of Ethanol produced:
According to equation;
180.156 g (1 mole) C₆H₁₂O₆ produces = 92.14 g (2 moles) of C₂H₅OH
So,
200 g of C₆H₁₂O₆ will produces = X g of C₂H₅OH
Solving for X,
X = (200 g × 92.14 g) ÷ 180.156 g
X = 102.29 g of C₂H₅OH
Mass of Carbon Dioxide produced:
According to equation;
180.156 g (1 mole) C₆H₁₂O₆ produces = 88.02 g (2 moles) of CO₂
So,
200 g of C₆H₁₂O₆ will produces = X g of CO₂
Solving for X,
X = (200 g × 88.02 g) ÷ 180.156 g
X = 97.71 g of CO₂
Answer:
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Answer:
-285.4 J/K
Explanation:
Let's consider the following balanced equation.
HCl(g) + NH₃(g) ⇒ NH₄Cl(s)
We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.
ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))
ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol
ΔS°r = -285.4 J/K