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trasher [3.6K]
3 years ago
10

Tetrachloroethane is a valuable nonflammable solvent. It's percent composition is 14.31% carbon, 1.20% hydrogen, and 84.49% chlo

rine. What is the empirical formula of this compound?
Chemistry
1 answer:
loris [4]3 years ago
6 0
First take all percents and make them grams. Since you're not given a overall molar mass you can assume it is 100 and therefore the percents are their masses.
So you have 14.31g Carbon, 1.2g Hydrogen, and 84.49g of Chlorine. Next you divide each by their molar masses to get moles of each.

Carbon= <u>14.31</u>g       Hydrogen= <u>1.2</u>g         Chlorine= <u>85.49</u>g
               12.01g                         1.01g                         35.45g
          = 1.19moles                = 1.188moles          = 2.411moles

Next you divide each of those numbers by the smallest, in this case, Hydrogen.
Thus, 
    Carbon= <u>1.19moles</u>     Hydrogen= <u>1.188moles</u>   Chlorine= <u>2.411moles</u>
                   1.188moles                      1.188moles                   1.188moles 
             =1.002                               =1                                =2.02
These are all close enough to round, so your final empirical formula is: CHCl2
Hope that helps!!
 



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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0
3 years ago
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