State the colour of a solution of copper sulphate

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0

3 years ago

An automobile is driving uphill. Which form of energy is not involved in this process?

Lapatulllka [165]

The answer is D) Electromagnetic energy. This is not a type of energy that occurs when a vehicle is moving.

7 0

3 years ago

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depr

lutik1710 [3]

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Using the equation,

Δ $T_{f}$ = i $K_{f}$ m

where:

Δ $T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ $T_{f}$ = i $K_{f}$ m we get,

Δ $T_{f}$ = 2 × 1.86 × 0.1219

Δ $T_{f}$ = 0.4536°C

So, Δ $T_{f}$ of solution is,

Δ $T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ $T_{f_{solution} }$ = - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C

Learn more about freezing point here;

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7 0

1 year ago

Hydrochloric acid reacts with sodium hydroxide to form water and sodium chloride. Hydrochloric acid is an extremely acidic, clea

Brut [27]

The answer is C.

You're welcome!

7 0

2 years ago

opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =

Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0

2 years ago