What is the chemical formula of the following compound

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

Balance the reaction for the combustion of pentane: ?C5H12+?O2→?CO2+?H2O Enter the four coefficients in order, separated by comm

Daniel [21]

Answer:

The four coefficients in order, separated by commas are 1, 8, 5, 6

Explanation:

We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.

C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O

We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.

As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.

Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.

We add 16 in the reactant side. Balanced equation is:

2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O

We also can divide by /2 in order to have the lowest stoichiometry

C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O

6 0

3 years ago

(( PLEASE HELP QUICK ))Which expression can be used to calculate the ratio of the rate of effusion of gas A to the rate of effus

borishaifa [10]

Answer:

The square root of the molar mass of B ÷ the square root of the molar mass of A

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

$r \propto \dfrac{1}{\sqrt{M}}$

If you have two gases A and B, the ratio of their rates of effusion is

$\dfrac{r_{\text{A}}}{r_{\text{B}}} = \sqrt{\dfrac{M_{\text{B}}}{M_{\text{A}}}}$

3 0

3 years ago

Give an example of competition in an ecosystem

neonofarm [45]

hbzjaj

Explanation:

competition for food

5 0

3 years ago

Which of these descriptions are properties of salts? Check all that apply.

Stels [109]

❌formed by 2 ions

❌ionic bonds

✅yes because of solvation

✅yes electrolytes are substances that break apart into ions when dissolved

✅in itself but salt water is a good conductor

✅such as water

4 0

3 years ago

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