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3 years ago

How many molecules are there in 2.30g of NH3

Chemistry

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

8.13x10^22 molecules

Explanation:

We can use the Avogadro's number(6.022 x 10^23 units / mole)

2.30 g NH3 (1 mol / 17.03 g ) (6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules

Hope this helps! Feel free to ask any questions!

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Answer:

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Explanation:

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4 years ago

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A disadvantage of sexual reproduction is that

sineoko [7]

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3 years ago

A student needed to dissolve a substance that she knew was soluble in water. according to the chart, which other solvent would m

Dmitry [639]

The chart is attached below and the options are as follow,

A Benzene

B Methanol

C Hexane

D Octane

Answer:
Option-B (Methanol) is the correct answer.

Explanation:
As we know that the substance given to student is soluble in water. Therefore, those substances which are soluble in water are polar in nature because water is polar in nature and Like Dissolves Like.
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3 years ago

How many milliliters of 0.183 m hcl would be required to titrate 5.93 g koh?

jek_recluse [69]

Molar mass:
KOH = 56.0 g/mol

Number of moles of KOH :

5.93 / 36.5 => 0.1624 moles

<span>KOH + HCl = KCl + H₂O
</span>
1 mole KOH --------------> 1 mole HCl
0.1624 moles KOH ----> ?

moles HCl = 0.1624 x 1 / 1

= 0.1624 moles of HCl

V ( HCl ) = moles / molarity

V(HCl) = 0.1624 / 0.183

V (HCl) = 0.887 L x 1000 = 887 mL

hope this helps!

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3 years ago

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

4 years ago