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katrin [286]
2 years ago
11

A student prepares two solutions as shown. The dots represent solute particles. The student wants to test the conductivity of ea

ch solution. Prior to carrying out the investigation, the student needs to identify the variables being controlled and the variables being changed between the two solutions.
A.
The volume of the solution and the number of solute particles are being changed between the two solutions, but the concentration of the solution is being held constant.

B.
The number of solute particles and the concentration of the solution are being changed between the two solutions, but the volume is being held constant.

C.
The number of solute particles is being changed between the two solutions, but the volume and concentration of the solution is being held constant.

D.
The volume of the solution and the concentration of the solution are being changed between the two solutions, but the number of solute particles is being held constant.
Chemistry
1 answer:
stealth61 [152]2 years ago
3 0

Answer:

I think its D

Explanation:

It cant be B or C bc the solute particles arent changing. And  its not A because the concentration isnt constant

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What is the molarity (M) of the following solutions?
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Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

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Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

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