Answer:
B. alanine aminotransferase.
Explanation:
Ketogenic amino acid contain low carbohydrate and have the keto acid group present in them. Examples of ketogenic amino acid include alanine, cysteine, glycine, serine, and threonine. These amino acids are usually converted to pyruvate.
Alanine aminotransferase helps by catalyzing the transfer of an amino group from alanine to alpha-ketoglutarate in the alanine cycle to form pyruvate and glutamate.
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.
Answer:
- A) pH = 2.42
- B) pH = 12.00
Explanation:
<em>The dissolution of HCl is HCl → H⁺ + Cl⁻</em>
- To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.
The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:
= 9.60*10⁻³ mol H⁺
So the concentration of H⁺ is
[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M
pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42
- <em>The dissolution of NaOH is NaOH → Na⁺ + OH⁻</em>
- Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:
= 0.02 mol OH⁻
[OH⁻] = 0.02 mol / 2.0 L = 0.01
pOH = -log [OH⁻] = -log (0.01) = 2.00
With the pOH, we can calculate the pH:
pH + pOH = 14.00
pH + 2.00 = 14.00
pH = 12.00
Answer:
Hg∧2+ has a negative standard entropy because the ions are highly solvated in aqueous phase; smaller the ionic size, the more highly it is surrounded by solvated ions. Therefore it will be in highly ordered state hence the entropy decreases.
Hg2 ^2+ has a positive standard entropy because the ionic size of Hg^2+ is smaller than Hg2 ^2+, so therefore the Hg^2+ is highly solvated and that means that it is in highly ordered state. Hg2 ^2+ is not highly solvated so it will have a positive entropy
Explanation:
The values of standard entropy of aqueous ions has a negative standard entropy because the ions are highly solvated in aqueous phase; smaller the ionic size, the more highly it is surrounded by solvated ions. Therefore it will be in highly ordered state hence the entropy decreases.
Hg2 ^2+ has a positive standard entropy because the ionic size of Hg^2+ is smaller than Hg2 ^2+, so therefore the Hg^2+ is highly solvated and that means that it is in highly ordered state. Hg2 ^2+ is not highly solvated so it will have a positive entropy
POH+pH=14
pH=14-pOH=14-3.53=10.47
pH=-log[H⁺]
[H⁺]=10^(-pH)
[H⁺]=10^(-10.47)=3.4*10(^-11)
