What substances is the most soluble in water?

Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate.

lawyer [7]

Answer:

Keq = 0.053

7.3 kJ/mol

Explanation:

Let's consider the following isomerization reaction.

glucose 6‑phosphate ⇄ glucose 1 - phosphate

The concentrations at equilibrium are:

[G6P] = 0.19 M

[G1P] = 0.01 M

The concentration equilibrium constant (Keq) is:

Keq = [G1P] / [G6P]

Keq = 0.01 / 0.19

Keq = 0.053

We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.

ΔG° = -R × T × lnKeq

ΔG° = -8.314 J/mol.K × 298 K × ln0.053

ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol

7 0

3 years ago

Which reaction is an example of an acid-base reaction? Which reaction is an example of an acid-base reaction?

oksian1 [2.3K]

Answer:

B) H2SO4 (aq) + Ca(OH)2 (aq) → CaSO4 (aq) + 2 H2O(l)

Explanation:

A is a reaction between a salt FeCl3 and a base KOH

C is a n acid decomposing on it's own to form two products

D is mercury, a metal reacting with oxygen. Two elements reacting. Neither are an acid or a base

E. is an acid reacting with a metal to liberate hydrogen. There is no base

3 0

3 years ago

When a solid with a mass of 53.78 g is added to 50.0 ml of water in a graduated cylinder, the volume increases to 56.4 ml. Calcu

Bad White [126]

Answer:

<h3>The answer is 8.40 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

volume = final volume of water - initial volume of water

volume = 56.4 - 50 = 6.4 mL

We have

$density = \frac{53.78}{6.4} \\ = 8.403125$

We have the final answer as

Hope this helps you

3 0

2 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

The name given to the Earth's climate when there was no ice at the poles

Citrus2011 [14]

Answer:

i think snowball, it sounds weird but its true (i think im sorry if its wrong)

Explanation:

7 0

3 years ago