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3 years ago

What is meant by “normal” when you are asked for normal boiling and melting points?

Chemistry

1 answer:

RideAnS [48]3 years ago

8 0

Answer:

Normal boiling point is the temperature at which a liquid boils at 1 atmosphere of pressure.

Explanation:

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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

How is the ionization energy, E, related to a period of elements?

Rainbow [258]

The valence electrons are further from the nucleus when moving from left to right

5 0

4 years ago

Read 2 more answers

What is a supercontinent? <br><br> please help me!

oee [108]

Answer:

earth land assembling together to form single large landmass

7 0

3 years ago

KF is a strong electrolyte, & HF is a weak electrolyte. How are they different?

Vitek1552 [10]

The strong electrolyte KF completely dissociates into K+ and F- ions when it dissolves in water. When the weak electrolyte HF dissolves in water, there are a few ions of H+ and F- present, but mostly dissolved HF molecules. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

7 0

3 years ago

4FeS2 + 11O2 > 2Fe2O3 + 8SO2

BartSMP [9]

Answer:

80.13g of SO2

Explanation:

4FeS2 + 11O2 —> 2Fe2O3 + 8SO2

Molar Mass of FeS2 = 55.8 + (2x32) = 119.8g/mol

Mass of FeS2 from the balanced equation = 4 x 119.8 = 479.2

Molar Mass of SO2 = 32 + (16x2) = 32 + 32 = 64g/mol

Mass of SO2 from the balanced equation = 8 x 64 = 512g

From the equation,

479.2g of FeS2 produced 512g of SO2.

Therefore, 75g of FeS2 will produce = (75 x 512)/479.2 = 80.13g of SO2

3 0

4 years ago