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bearhunter [10]
3 years ago
14

A certain metal M forms a soluble nitrate salt M(NO3), Suppose the left half cell ofa galvanic cell apparatus is filled with a 4

.00 M solution of M (NO,), and the right half cell with a 20.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show?
Chemistry
1 answer:
kozerog [31]3 years ago
6 0

Answer:

Explanation:

The left electrode will be positive on the grounds that focus on the concentration that the cell electron moves from a lower concentration fixation to a higher concentration. Thus right electrode will go about and act as an anode and will be negative.  Also, the left electrode will be the cathode and will be positive.

The concentration cell E_{Cell} = \dfrac{-0.0591}{1} log \dfrac{0.02}{4}

E_{Cell} =0.136 V

         = 136 mV

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Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur
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Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of C_5H_{12} = 21.9 g

Molar mass of C_5H_{12} = 72.15 g/mole

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First we have to calculate the moles of C_5H_{12}.

\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles

Now we have to calculate the moles of H_2O.

The balanced chemical reaction will be,

C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)

From the balanced reaction we conclude that

As, 1 mole of C_5H_{12} react to give 6 moles of H_2O

So, 0.3035 moles of C_5H_{12} react to give 0.3035\times 6=1.821 moles of H_2O

Now we have to calculate the mass of H_2O.

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g

Therefore, the mass of water produced will be 32.78 grams.

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