Determine the number of moles in 1.5 x 10^25 atoms of iron.

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

Explain why atoms would gain or lose electrons in order to resemble the elements in the particular group you named above. Consid

posledela

Answer:

The number of electrons in the outermost shell of an element is represented in the periodic table as the group number that element is situated in. ... the number of electrons in all shells of an element is represented in the periodic table as the element's atomic number.

Explanation:

thatdummyemily

hope this helps srry if it doesn't tho, but if it does help can u plz mark me brainlist

6 0

2 years ago

water and ammonia are polar. The hydrogen atoms in each have a slight positive charge. What does that tell you about the oxygen

Andreas93 [3]

Answer:

More electronegative

Explanation:

The oxygen and nitrogen in ammonia shows that they are more electronegative than the hydrogen atoms.

Electronegativity shows the affinity of an atom for valence electrons.
Electronegativity is the measure of the relative tendency with which the atoms of the element attract valence electrons in a chemical bond.

The oxygen atom in water and the nitrogen atom in ammonia are more electronegative.

They attract the electrons more and they then become negatively charged.

The hydrogen will then become positively charged in the shared covalency.

4 0

3 years ago

A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

Answer: The freezing point of the solution is $-17.6^oC$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $0.80^oC$

Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

6 0

3 years ago

A scientist adds iodine as an indicator to an unknown substance. What will this indicator reveal about the substance?

VLD [36.1K]

Answer:

it will turn blue black to any substance

7 0

2 years ago