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notsponge [240]
3 years ago
15

How many mol of Cl are in 10.0 g of Cl? The molar mass of Cl is 35.45 g/mol

Chemistry
2 answers:
Delvig [45]3 years ago
6 0

Answer:

\boxed {\boxed {\sf 0.282 \ mol \ Cl}}

Explanation:

The molar mass, or mass of 1 mole of a substance, is used to convert grams to moles. We are given the molar mass of chlorine. It is 35.45 grams per mole.

We convert using dimensional analysis, so we must set up a conversion factor using the molar mass.

\frac { 35.45 \ g \ Cl}{1 \ mol \ Cl}

We are converting 10.0 grams of chlorine to moles, so we must multiply the conversion factor by this value.

10.0 \ g  \ Cl* \frac { 35.45 \ g \ Cl}{1 \ mol \ Cl}

Flip the conversion factor. The result is equivalent and allows us to cancel the units of grams of chlorine.

10.0 \ g  \ Cl* \frac {1 \ mol \ Cl}{ 35.45 \ g \ Cl}

10.0 * \frac {1 \ mol \ Cl}{ 35.45 }

\frac {10.0 }{35.45} \ mol \ Cl

0.282087447 \ mol \ Cl

The original value of grams (10.0) has 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 0 in the ten-thousandth place tells us to leave the 2 in the thousandth place.

0.282 \ mol \ Cl

Approximately <u>0.282 moles of chlorine</u> are in 10.0 grams of chlorine.

postnew [5]3 years ago
5 0
There are 3.545 mol in 10 g of CL
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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ
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Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

<u>Step 1:</u> Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

<u>Step 2:</u> Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+  in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed  in the back titration with Ca2+:

<u>Step 3: </u>Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

<u>Step 4:</u> Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

<u>Step 5:</u> Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction  with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

<u>Step 6:</u> Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

<u>Step 7: </u>Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

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