It is known that for high concentration of $M^{2+}$ , reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, $E^{o}_{cell}$ = 0 and the general reaction equation is as follows.

$M^{2+} + M \rightarrow M + M^{2+}$

3.00 M n = 2 30 mM

E = $0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}$

= $-\frac{0.0591}{2} log (5 \times 10^{-2})$

= 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

5 0

3 years ago

What is the identity of the element whose isotopes you have selected

Pachacha [2.7K]

Answer:

Rubidium

Explanation:

Nil

5 0

2 years ago

Describe how stirring, surface area and temperature affect the rate of dissolving.

denis23 [38]

Answer:

friction

Explanation:

since it has a high tempature the friction increases like blowing air in a furnace

7 0

1 year ago

A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.

MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0

2 years ago