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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 a balloon with volume V. It contains n moles of gas and has an internal pressure of P. The temperature of the gas is T.
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Answer:

V₂ = 0.6 V.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n is constant, and have different values of P, V and T:

<em>(P₁V₁T₂) = (P₂V₂T₁).</em>

<em></em>

V₁ = V, P₁ = P, T₁ = T.

V₂ = ??? V, ​P₂ = 1.25 P, T₂ = 0.75 T.

<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>

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3 years ago
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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
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Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

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Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

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Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

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the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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