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aleksley [76]
3 years ago
14

Ni-cad (nickel–cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 v normally used in

flashlights. they have a cadmium anode, which is oxidized to cd(oh)2(s) in a basic background electrolyte of concentrated oh–, and a cathode half-reaction in which nio(oh)(s) is reduced to ni(oh)2(s). what is the value of δg°cell?
Chemistry
1 answer:
Lorico [155]3 years ago
6 0
The half-reaction are:

Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.

NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.

Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.

ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>


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In which 2 states of matter can a substance take the shape of its container?
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When 0.25 moles of NH3 are produced according to the reaction below, how many moles of H2 must have reacted?
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nNH3/2=nH2/3

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Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl. NH3(g)+HCl(g)⟶NH4
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9.41 atm is the pressure in atmospheres of the gas remaining in the flask

<h3>What is the pressure in atmospheres?</h3>

The equation NH3(g) + HCl(g) ==> NH4Cl(s) is balanced.

Divide the moles of each reactant by its coefficient in the balanced equation, and the limiting reagent is identified as the one whose value is less. With the issue we now have...

6.44 g NH3 times 1 mol NH3/17 g equals 0.3688 moles of NH3 ( 1 = 0.3688)

HCl: 6.44 g of HCl times one mole of HCl every 36.5 g equals 0.1764 moles ( 1 = 0.1764). CONTROLLING REAGENT

NH4Cl will this reaction produce in grams

0.1764 moles of HCl multiplied by one mole of NH4Cl per mole of HCl results in 9.44 g of NH4Cl (3 sig. figs.)

the gas pressure, measured in atmospheres, that is still in the flask

NH3(g) plus HCl(g) results in NH4Cl (s)

0.3688......0.1764............0..........

Initial

-0.1764....-0.1764........+0.1764...Change

Equilibrium: 0.1924.......0...............+0.1924

There are 0.1924 moles of NH3 and no other gases in the flask. This is at a temperature of 25 °C (+273 = 298 °K) in a volume of 0.5 L. After that, we may determine the pressure by using the ideal gas law (P).

PV = nRT

P = nRT/V = 0.1924 mol, 0.0821 latm/mol, and 298 Kmol / 0.5 L

P = 9.41 atm

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

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An allele whose trait shows up in an organism when the allele is present is a
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