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Alexeev081 [22]

3 years ago

7

Describe the best way Ryan can increase the rate of dissolving of lead (II) chloride in a solution.

2 answers:

eduard3 years ago

8 0

Answer: A) Increase the temperature of water the water

pychu [463]3 years ago

4 0

Solubility is the amount of solute that can dissolve in a given solvent at a specified temperature or pressure. Therefore, solubility is affected by pressure and temperature such that increase in temperature increases the rate of solubility of substances. In additional solubility depends on the molecular size of the solute such that when other factors are kept constant then a solute with small particles will dissolve more faster. Therefore, to increase the rate of solubility Ryan may; increase the temperature of water.

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What is the mass in grams of 1.204 x 1024 molecules of

natulia [17]

The answer is C) 4.03 g

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3 years ago

Question 1 Points 3 23 and Louis immerses his left hand in a beaker containing cold water and immerses his right hand in a beake

olganol [36]

Answer:look down below

Explanation:

6 0

3 years ago

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon decomposes after 5.55 days if the sample i

Leni [432]

Answer:

The mass of radon that decompose = 63. 4 g

Explanation:

R.R = P.E/(2ᵇ/ⁿ)

Where R.R = radioactive remain, P.E = parent element, b = Time, n = half life.

Where P.E = 100 g , b = 5.55 days, n = 3.823 days.

∴ R.R = 100/ $2^{5.55/3.823}$

R.R = 100/ $2^{1.45}$

R.R = 100/2.73

R.R = 36.63 g.

The mass of radon that decompose = Initial mass of radon - Remaining mass of radon after radioactivity.

Mass of radon that decompose = 100 - 36.63

= 63.37 ≈ 63.4 g

The mass of radon that decompose = 63. 4 g

8 0

3 years ago

Why do people write fake answers that just restate the question, get the points for answering, then the moderation team does not

SVETLANKA909090 [29]

I really don't know and it sucks when i need help and people are giving me false answers.

7 0

3 years ago

Read 2 more answers

In nature, one common strategy to make thermodynamically unfavorable reactions proceed is to couple them chemically to reactions

maks197457 [2]

Answer:

31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.

Explanation:

A⇌B, $\Delta G_1$ = 11.9 kJ/mol ...[1]

B⇌C, $\Delta G_2$ = -26.7 kJ/mol ...[2]

C⇌D, $\Delta G_3$ = 7.30 kJ/mol ...[3]

A⇌D, $\Delta G_4$ = ?...[4]

[4] = [1] - [2] - [3] (Using Hess's law)

$\Delta G_4=\Delta G_1-(\Delta G_2+\Delta G_3)$

$=11.9 kJ/mol - (-26.7 kJ/mol+7.30 kJ/mol)$

$=\Delta G_4=31.3 kJ/mol$

31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.

7 0

4 years ago