Bronsted lowry bases
NO2- amd OH-
Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
Answer:
e
Explanation:
<em>Provided the reaction that leads to the formation of the products can proceed in both forward and backward directions, the correct answer would be yes because the reaction will proceed backward until equilibrium is reached.</em>
<u>For a reaction that can proceed both forward and backward, the addition of a catalyst increases the rate of reaction in both directions based on the fact that a catalyst cannot alter the equilibrium of a reaction. </u>
Hence, if an enzyme is added to the product of a reaction that has the potential to proceed in both forward and reverse reactions, a substrate would be expected to form because the reaction will proceed backward until an equilibrium is reached.
The correct option is e.
Answer:
a. 9.2
b. 4.4
c. 6.3
Explanation:
In order to calculate the pH of each solution, we will use the definition of pH.
pH = -log [H⁺]
(a) [H⁺] = 5.4 × 10⁻¹⁰ M
pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2
Since pH > 7, the solution is basic.
(b) [H⁺] = 4.3 × 10⁻⁵ M
pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4
Since pH < 7, the solution is acid.
(c) [H⁺] = 5.4 × 10⁻⁷ M
pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3
Since pH < 7, the solution is acid.
