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Anna007 [38]
3 years ago
13

Five different substances are given to you to be dissolved in water. Which substances are most likely to undergo dissolution in

water?lithium iodide, LiIsodium fluoride, NaFpotassium bromide, KBrbenzene, C6H6hexane, C6H14
Chemistry
1 answer:
9966 [12]3 years ago
7 0

Answer:

NaF and KBr

Explanation:

These two substances contain pure ionic bonds. LiI is a covalent compound because of the Large polarizing power of Li and the high polarizability of I(Fajan's rules). The other compounds mentioned are organic compounds. They are not soluble in water at all.

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Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as
Sonbull [250]

Answer:

0.100 m AlCl3  will have the highest boiling point

Explanation:

Step 1: Data given

The molal boiling point elevation constant for water is 0.51°C/m

Since those are all  aqueous solutions, the have the same molal boiling point elevation constant

Step 2:

0.100 m C6H12O6

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = 1

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT = 1 * 0.51 * 0.100

ΔT  = 0.051 °C

0.100 m NaCl

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Na+ + Cl- = 2

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =2 * 0.51 * 0.100

ΔT = 0.102 °C

0.100 m AlCl3

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Al^3+ + 3Cl- = 4

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =4 * 0.51 * 0.100

ΔT = 0.204 °C

0.100 m MgCl2

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Mg^2+ +2Cl- = 3

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =3 * 0.51 * 0.100

ΔT = 0.153 °C

0.100 m AlCl3  will have the highest boiling point

5 0
3 years ago
A chemist prepares a solution of copper(l) sulfate (CuSo4) by measuring out 11.7 g of copper(II) sulfate into a 350. mL volumetr
Ksju [112]

Answer:

0.209 mol/L

Explanation:

Given data

  • Mass of copper(lI) sulfate (solute): 11.7 g
  • Volume of solution: 350 mL = 0.350 L

The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:

11.7 g × (1 mol/159.61 g) = 0.0733 mol

The molarity of copper(Il) sulfate is:

M = moles of solute / liters of solution

M = 0.0733 mol / 0.350 L

M = 0.209 mol/L

6 0
3 years ago
For maleic acid, hoocch=chcooh, ka1 = 1.42  10–2 and ka2 = 8.57  10–7 . what is the concentration of maleate ion (–oocch=chcoo
uranmaximum [27]

ANSWER

Find the attachment

7 0
3 years ago
Read 2 more answers
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