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sveticcg [70]
3 years ago
9

2. What ions are present in what ratio in a solution of aqueous calcium chloride?

Chemistry
1 answer:
Alenkasestr [34]3 years ago
5 0

Answer:

\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form \mathrm{Ca}^{2+}, while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two\mathrm{Cl}^{-} ions.

\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one \mathrm{Ca}^{2+}and two\mathrm{Cl}^{-}ions: \mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})

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The relative molecular mass of the gas​ : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

The relative molecular mass of the gas​

Solution

Graham's Law

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas​

Input the value :

\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol

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Which of these factors will cause a solid solute to dissolve faster?
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Read 2 more answers
Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • <u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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Group 7a would have an ion charge of -1 because it has 7 valence electrons and it wants to gain one more electron(which is negative) to have a full shell of 8
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