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vivado [14]
3 years ago
14

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L
Chemistry
1 answer:
BaLLatris [955]3 years ago
4 0

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

Explanation :

As we know that CaCO₃ dissociates to give Ca^{2+} ion and CO_3^{2-} ion.

The solubility equilibrium reaction will be:

CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}

The expression for solubility constant for this reaction will be,

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

Let solubility of CaCO₃ be, 's'

K_{sp}=(s)\times (s)

K_{sp}=s^2

8.7\times 10^{-9}=s^2

s=9.33\times 10^{-5}mol/L

Therefore, the solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

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The number of electrons in the outermost principal energy level of a chlorine atom is
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seven electrons

Explanation:

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Abbreviated electronic configuration:

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2. its melting point is 172.2 K

3. its boiling point is 238.6 K

4. it is disinfectant and can kill the bacteria.

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3 years ago
Avogrados number of representatives particles is equal to one what?​
kotykmax [81]

Answer:

Avogadro number of representatives particles is equal to one mole.

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For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

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Answer: 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles

The balanced chemical equation is:

NH_4NO_3\rightarrow N_2O+2H_2O  

According to stoichiometry :

1 mole of NH_4NO_3 produce = 1 mole of N_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{1}{1}\times 0.571=0.571moles  of N_2O  

Mass of N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g

1 mole of NH_4NO_3 produce = 2 moles of H_2O

Thus 0.571 moles of NH_4NO_3 will require=\frac{2}{1}\times 0.571=1.142moles  of H_2O  

Mass of H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g

Thus 25.13 g of N_2O  and 20.56 g of H_2O will be produced from 45.70 g of NH_4NO_3

5 0
3 years ago
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