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Art [367]
4 years ago
9

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCl. What volume of 3.00 M HCl did the stu

dent use to make the sample? Use mc020-1.jpg. 3.70 mL 16.7 mL 30.0 mL 83.3 mL
Chemistry
2 answers:
laila [671]4 years ago
7 0
A) 3.70 mL   B)   16.7 mL   C) 30.0 mL   D) 83.3 mL
<span>
It is C 30.0 mL


</span>
vlabodo [156]4 years ago
3 0

Answer is 30.0mL

according to the question dilution is performed by the student which can be calculated with the help of :

                          M_{1}V_{1}=M_{2}V_{2}     EQUATION (1)


Here, we have:

 M_{1} = 3M

  V_{1}  = y mL

M_{2}= 1.80 M

V_{2} = 50.0mL

putting the known values in equation (1) which will give us value of V_{1}

                                         3.00× y =1.80 ×50.0

                                                         y= 30.0mL

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4 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
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Answer:

3.47 ×10^-10

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The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

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