Can someone help quick

A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of

castortr0y [4]

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains $1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$ .

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$ .

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.

6 0

3 years ago

Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml.

OleMash [197]

Answer: -

15.55 M

35.325 molal

Explanation: -

Let the volume of the solution be 1000 mL.

Density of nitric acid = 1.42 g/ mL

Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid

= 1000 mL x 1.42 g/ mL

= 1420 g.

Percentage of HNO₃ = 69%

Amount of HNO₃ = $\frac{69} {100} x 1420 g$

= 979.8 g

Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol

Number of moles of HNO₃ = $\frac{979.8 g}{63 g/ mol}$

= 15.55 mol

Molarity is defined as number of moles per 1000 mL

We had taken 1000 mL as volume and found it to contain 15.55 moles.

Molarity of HNO₃ = 15.55 M

Mass of water = Total mass of nitric acid solution - mass of nitric acid

= 1420 - 979.8

= 440.2 g

So we see that 440.2 g of water contains 15.55 moles of HNO₃

Molality is defined as number of moles of HNO₃ present per 1000 g of water.

Molality of HNO₃ = $\frac{15.55 x 1000}{440.2}$

= 35.325 molal

3 0

3 years ago

Mixture of same boiling point cannot be separated by using distillation why?

Bond [772]

Answer:

both the substances will evaporate

Explanation:

3 0

2 years ago

Read 2 more answers

Predict the products of each of these reactions and write balanced complete ionic and net ionic equations for each. If no reacti

Bumek [7]

Answer:

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq) → AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq) + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) + Li⁻(aq) + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq) → Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq) + SO²⁻₄(aq) + 2Li⁺(aq) + SO₃²⁻(aq) → 2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq) → SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq) → Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq) → Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq) → 2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq) → LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq) → 3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq) → 3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) + 3OH⁻ (aq) → Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq) → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq) → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq) → Hg₂Cl₂ (s)

8 0

2 years ago

The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg

almond37 [142]

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

$pH = \frac{1}{2} (9,1+10,5) =$ 9,8

For arginine:

$pH = \frac{1}{2} (9,0+12,5) =$ 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, pH 9,8 is likely to work best for this separation

I hope it helps!

5 0

3 years ago