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3 years ago

What is the oxidation state of an individual phosphorus atom in P O 3 3−?

Chemistry

1 answer:

Tju [1.3M]3 years ago

4 0

-3 - 3(-2) = +3
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Two positively charged objects (N pole) are separated by a large distance. One of the positively charged objects (N pole) is rep

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Answer:

Explanation:

Applying law of electrostatic which states that like charges repel each other and unlike charges attract each other

N and S are unlike charges that turn and make the former repulsive force (due to two like charges N and N)to reduce and attractive force between N and S to increase

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4 years ago

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3 years ago

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Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

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3 years ago