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Inessa [10]
3 years ago
14

Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the ne

gative of the measured value if the polarity is not the same as the standard cell.) Take [Cu2+] to be 1 M. Show your calculations. Ecell=0.4249v, Ecell=-0.00191. M E =E- 0.0592/2 • log ([Cu2+]/[Ag+1?) (10) cell
Chemistry
1 answer:
mixer [17]3 years ago
5 0

Answer:

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

<em>Where E</em>⁰<em>= 0.4249V</em>

<em>E(Cell) = -(-0.0019V) -Measured value-</em>

<em>[Cu2+] = 1M</em>

<em />

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

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how do the boiling point and freezing point of a solution of water and calcium chloride at standard pressure compare to the boil
Snezhnost [94]
Both the increase in the boling point and the depression on the freezing point are colliative properties.

This is, they are proportional to the number of particles dissolved in the solvent, which is measured by the molality of the solution and the factor i (Van'f Hoff).

The answer to the question is that 1) the boling point of a solution of water and calcium chloride at standard pressure will be higher than the normal boiling point of pure water, and 2) the freezing point of a solution of water and calcium chloride at standard pressure will be lower than the normal freezing point of pure water.
3 0
3 years ago
25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is
vlabodo [156]

Answer:

a)   119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

\bigtriangleup T_f=K_fm, \  \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\

#We use the molality above to calculate the molar mass:

m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol

Hence, the molar mass of the compound is 119 g/mol

5 0
3 years ago
Using your periodic table, determine the electron configuration of Barium ion, Ba+2
seraphim [82]

Answer:

im pretty sure the electron configuration of the barium ion Ba+2 is [Kr]5s24d105p6

Explanation:

4 0
3 years ago
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inysia [295]

Answer:

over 10 weeks the plant will grow 24 time longer than the the same hieght of the plant and the plant age will become 45 and so the plant is very much give hey man is my answer corecct

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3 years ago
Read 2 more answers
Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).
likoan [24]

Answer:

1.40*10⁻² M

Explanation:

We have the solubility formula

Solubility,

S = KH*P  

where

KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm

P = atmospheric pressure = 0.400 atm

Hence, we have

S = KH*P

= (3.50*10⁻² mol/L.atm)*(0.400 atm)

= 1.40*10⁻² mol/L

But 1 mol/L = 1 M,

Hence, the answer (1.40*10⁻² mol/L ) is equivalent to

= 1.40*10⁻² M

5 0
3 years ago
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