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3 years ago

Indicate what happens to the concentration of Pb2+ in each half cell

Chemistry

1 answer:

Taya2010 [7]3 years ago

3 0

Answer:

Your questions requires diagrams of the cell to get which one is on the left or right. However, see the attached file below

The correct answer is (d) the left half-cell will decrease in concentration; and the right half-cell will increase in concentration.

Explanation:

The concentration of the Pb2+ increases in the oxidation half cell while the concentration of the Pb2+ decreases in the reduction half cell during the reaction.

In the Left Beaker (Left half cell), their is less concentration

Pb(s) ---> Pb2+(aq) + 2 e- Concentration of Pb2+(aq) increase ; Electrons going out from this side

In the Right Beaker (right half cell), their is more concentration

Pb2+(aq) + 2 e- ---> Pb(s) Concentration of Pb2+(aq) decrease ; Electrons coming in to this side

Electrons will flow from Left to Right direction.

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2 years ago

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Answer:

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Explanation:

5 + 3.3 = 8.5

But you should use sig figs.

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3 years ago

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3 years ago

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The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti

lisabon 2012 [21]

Answer:

Option E, Half life = $2.96\times 10^3\ s$

Explanation:

For a first order reaction, rate constant and half-life is related as:

$t_{1/2}=\frac{0.693}{k}$

Where,

$t_{1/2}$ = Half life

k = Rate constant

Rate constant given = $2.34\times 10^{-4}\ s^{-1}$

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4 0

3 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

Now put all the given values in the above formula, we get:

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

3 0

3 years ago