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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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A salt lover decided to flavor her meal by adding 2.50 moles of sodium chloride,
FromTheMoon [43]

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4 0
2 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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