15g of magnesium (II) oxide decomposes

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10-2M ) and calcium ion

Alona [7]

Answer : The remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, $1.07\times 10^{-6}M$

Explanation :

The dissociation of barium fluoride is written as:

$BaF_2\rightleftharpoons Ba^{2+}+2F^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ba^{2+}][F^-]^2$

As we know that at room temperature $25^oC$ the $K_{sp}$ of barium fluoride is, $1\times 10^{-6}$ .

Now put all the given values in this expression, we get:

$1\times 10^{-6}=(2.75\times 10^{-2})\times [F^-]^2$

$[F^-]=6.03\times 10^{-3}M=0.00603M$

The barium fluoride precipitate when fluoride ion is equal to 0.00603 M.

The dissociation of calcium fluoride is written as:

$CaF_2\rightleftharpoons Ca^{2+}+2F^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][F^-]^2$

As we know that at room temperature $25^oC$ the $K_{sp}$ of calcium fluoride is, $3.90\times 10^{-11}$ .

Now put all the given values in this expression, we get:

$3.90\times 10^{-11}=(6.70\times 10^{-2})\times [F^-]^2$

$[F^-]=2.41\times 10^{-5}M=0.0000241M$

The calcium fluoride precipitate when fluoride ion is equal to 0.0000241 M.

Since, the fluoride ion concentration in calcium fluoride is less then the fluoride ion concentration in barium fluoride. That means, calcium fluoride will precipitate first.

Thus, the concentration $Ca^{2+}$ ion remaining at $F^-$ concentration (0.00603 M) is calculated as:

$K_{sp}=[Ca^{2+}][F^-]^2$

Now put all the given values in this expression, we get:

$3.90\times 10^{-11}=[Ca^{2+}]\times (0.00603)^2$

$[Ca^{2+}]=1.07\times 10^{-6}M$

Therefore, the remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, $1.07\times 10^{-6}M$

3 0

3 years ago

126785033% Divided by 348675948849 = <br> what does it equal

meriva

Answer:

0.00000363618

could be wrong.

double check me someone or just trust me

(don't blame me if you get it wrong)

5 0

3 years ago

In a constant-pressure calorimetry experiment, a reaction gives off 22.8 kJ of heat. The calorimeter contains 150 g of water, in

ludmilkaskok [199]

Answer:

The final temperature of water is 56,7 ºC

Explanation:

We need to apply the specific heat capacity formula to solve this.

Q = m . C . ΔT

22,8 kJ = 150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

(We need to convert kJ in J to use the specific heat of water, be careful with units)

22,8 kJ = 22800 J (x1000)

22800 J = 150 g . 4,186 J/g °C . (Tfinal - 20,4º C)

22800 J = 627,9 J/ºC (Tfinal - 20,4º C)

22800 J / 627,9 ºC/ J = (Tfinal - 20,4º C)

36,3ºC = Tfinal - 20,4º C

36,3ºC + 20,4º C = 56,7 ºC

7 0

4 years ago

The parts of the Earth that contain living organisms

rodikova [14]

Answer:

Biosphere

The Biosphere --contains all the planet's living things. This sphere includes all of the microorganisms, plants, and animals of Earth. Within the biosphere, living things form ecological communities based on the physical surroundings of an area.

Explanation:

4 0

3 years ago

Read 2 more answers

What are three sources of information scientist use when they do research

exis [7]

Books, Other scientist's research, the Internet.

6 0

3 years ago