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brilliants [131]
4 years ago
10

What reagent could be used to separate Br- from NO3- when added to an aqueous solution containing both?

Chemistry
2 answers:
joja [24]4 years ago
4 0
<span>AgNO3(aq)
hope it helps
</span>
Phantasy [73]4 years ago
4 0

Answer:

The correct answer is silver nitrate solution that is AgNO_3(aq).

Explanation:

To separate Br^- from NO_{3}^- ions we will use silver nitrate solution. On addition of silver nitrate solution in a mixture of bromide and nitrate ions.

The silver ion will react with bromide ions to form silver bromide which will get precipitated out as yellow color solid from a mixture.

AgNO_3(aq)+Br^-(aq)\rightarrow AgBr(s)+NO_{3}^-(aq)

AgNO_3(aq)+NO_{3}^-(aq)\rightarrow no reaction

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If given the mole fraction of water in an aqueous sucrose (C12H22O11) solution and asked to calculate the molality of the soluti
olasank [31]

Answer:

No additional information is required in order to calculate the molality of the solution.

Explanation:

Mole fraction of water = \chi_1

Mole fraction of sucrose = \chi_2

\chi_1+\chi_2=1

From the given mole fraction of water mole fraction of sucrose can be determined.

And from both values of mole fraction moles of water and sucrose can also be calculated.

\chi_1=\frac{n_1}{n_1+n_2}

\chi_2=\frac{n_2}{n_1+n_2}

After calculating the moles of water convert them into mass of water in grams and then change that into kilograms say that be M'.

Finally we can determine the molality of solution ;

Molality=\frac{\text{Moles of compound}}{\text{mass of solvent (kg)}}

Molality of the sucrose solution ;

m=\frac{n_1}{M'}

From this we can conclude that no additional information is required in order to calculate the molality of the solution.

7 0
3 years ago
Calculate the solubility at 25°c of cubr in pure water and in a 0.0030m cobr2 solution. you'll find ksp data in the aleks data t
Reil [10]

Answer:

the solubility at 25°c of cubr in pure water = 0.011 g/L

The solubility at 25°c of cubr  in a 0.0030m cobr2 solution = 0.00014 g/L

Explanation:

Step 1: Data given

Ksp of CuBr = 6.27 × 10^-9

Molar mass CuBr = 143.45 g/mol

Step 2: Calculate the solubility at 25°c of cubr in pure water

Ksp = [Cu+][Br-]

The initial concentration

[Cu+] = 0M

[Br-] = 0M

The concentration at the equilibrium

[Cu+] = X M

[Br-] = X M

Ksp =6.27 * 10^-9 =  X * X = X²

S = 7.9 *10^-5 mol /l  

7.9 * 10^-5 mol/L * 143.45 g/mol = 0.011 g/L

Step 3: Calculate the solubility at 25°c of cubr  in a 0.0030m cobr2 solution.

The balanced equation:

CoBr2(aq) → Co^2+(aq) + 2 Br⁻(aq)

For 1 mol CoBr2 we'll have 1 mol Co^2+ and 2 moles Br-

For 0.0030 M CoBr2 we'll have 0.0030 M Co^2+ and 0.0060 M Br-

The initial concentration

[Cu+] = 0M

[Br-] = 0.0060 M

At the equilibrium

[Cu+] = X

[Br-] = 0.0060 + X

Ksp = 6.27 * 10^-9 = X * (0.0060+X)

6.27 *10^-9 = 0.0060 X + X²

X = 1.0 * 10^-6 mol/L

1.0 * 10^-6 mol/L * 143.45 g/mol = 0.00014 g/L

6 0
3 years ago
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