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nignag [31]
4 years ago
12

Which statement is true? CTP hydrolysis can be a source of chemical energy. dATP is the most common source of chemical energy in

cells. Hydrolysis of ATP to ADP and phosphate releases about 14 kJ/mol of free energy. Hydrolysis of the α, β and β, γ phosphoester bonds of a nucleotide is used as a source of chemical energy.
Physics
1 answer:
Gemiola [76]4 years ago
3 0

1. Statement A is true.

When GTP is hydrolysed, the free energy of hydrolyses is used to power or drive reactions that are favourable energetically.

2. Statement B it true.

ATP is a complex chemical that gives energy for the activities in many living cells. During hydrolyses, chemical energy stored in the energy-rich phosphoanhydride is released. Hence its a common source of chemical energy in cells.

3. Statement C is false.

The hydrolyses of ATP to ADP in the presence of phosphate, releases one mole of ATP which is estimated to be -57Kj/mol not 14Kj/mol. Below is the equation;

ATP + H20 -----> ADP + Pi + Free energy.

4. Statement D is false.

GTP stands for guanosine triphosphate.

ATP stands for adenosine triphosphate

ADP stands for adenosine diphosphate.

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the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
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Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

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