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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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<h3>What is the molar mass?</h3>

We know that;

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The molality of the solution is obtained from;

m = ΔT/K i

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How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602
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Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as 1.602 \times 10^{-19} C. Hence, we will calculate its number of moles as follows.

          Moles = \frac{mass}{\text{molar mass}}

                     = \frac{0.7 \times 1000 g}{7 g/mol}

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According to mole concept, there are 6.023 \times 10^{23} atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

        No. of atoms = 100 \times 6.023 \times 10^{23}

                               = 6.023 \times 10^{25} atoms

Since, it is given that charge on 1 atom is as follows.

                     3 \times 1.602 \times 10^{-19}C

                    = 4.806 \times 10^{-19}C

Therefore, charge present on 6.023 \times 10^{25} atoms will be calculated as follows.

    6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C

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Thus, we can conclude that a positive charge of 28.95 \times 10^{6}C is in 0.7 kg of lithium.

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