Chemistry, 06.07.2021 05:20 TheOneandOnly003

4.106 Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2

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Answer from: dianaosorio33895

$\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}$

$\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}$

Explanation:

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

$molarity= \frac {moles \ of \ solute}{liters \ of \ solution}$

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

${150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L$

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

$0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}$

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

$0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L$

$0.150 \ L *0.245 \ mol \ CaCl_2 /L=x$

The units of liters cancel.

$0.150 *0.245 \ mol \ CaCl_2 =x$

$0.03675 \ mol \ CaCl_2$

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

$\bold {0.0368 \ mol \ CaCl_2}$

2. Mass of the Solute

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Ca: 40.08 g/mol Cl: 35.45 g/mol

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Cl₂: 35.45 *2= 70.9 g/mol

Add calcium's molar mass.

CaCl₂: 40.08 + 70.9 =110.98 g/mol

Use the molar mass as a ratio.

$\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}$

Multiply the moles of calcium chloride we calculated above.

$0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}$

The units of moles of calcium chloride cancel.

$0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }$

$4.084064 \ g\ CaCl_2$

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

$\bold {4.08 \ g \ CaCl_2}$

Answer from: Quest

can you list the answers so i can answer

Answer from: Quest

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃ → 3 nicl₂ + 2 albarium and copper bromideba + cubr → no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn → znso₄ + na

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4.106 Calculate the moles and the mass of solute in each of the following solutions.
(a) 150....

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