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1 year ago

the deltag value for the acontinase reaction is 5.8 kj/mol. what is the ratio of isocitrate/citrate understandard conditions?

1 answer:

5 0

The ratio of isocitrate/citrate under standard conditions is 0.096 when the value of ΔG° is given by 5.8kj/mol.

We know that isocitrate and citrate are convertible. It is given that the ΔG° is given by 5.8kj/mol. Then we know that,

ΔG° = -RTlnK

lnK = -RT/ΔG°

On substituting the values, we get

lnK = -5.6x10³/(6.314x296)

lnK = -2.34

K = 0.09623

The value of k is 0.09623.

Therefore, the ratio of Isocitrate to citrate is given by,

[Isocitrate]/[Citrate] = 0.09623

The ratio could be however rounded off to 0.096.

Therefore, the ratio is given by 0.096 for the isocitrate/citrate under standard conditions. In this way, the ratio could be computed very easily.

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Which is an example of an exothermic process?

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3 years ago

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<u>Explanation:</u>

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Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g

Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

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