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lorasvet [3.4K]
3 years ago
14

What is the percentage composition of cucl

Chemistry
2 answers:
mart [117]3 years ago
6 0
The formula weight of CuCl is: 98.999 

<span>Percent compositions are usually weight percent (abbreviated w/o). </span>

<span>Since CuCl has one mole of each element, the calculation is simple. </span>

<span>For Cu it is: (63.546 / 98.999) x 100% = 64.189 w/o </span>

<span>For Cl it is 100.000 - 64.189 = 35.811 w/o </span>
valentina_108 [34]3 years ago
3 0

Answer:

Cu = 64.189%

Cl = 35.811%

Explanation:

Percentage composition = (mass/molar mass) x 100

CuCl Molar Mass 98.999g/mol

Cu = 63.546g/mol

Cl = 35.453g/mol

Cu = (63.546/98.999) x 100 = 64.189

Cl =  (35.453/98.999) x 100 = 35.811

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
2 years ago
46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride
slega [8]

Answer:

m_{HgCl_2}=42.7gHgCl_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2

Best regards.

3 0
2 years ago
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