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saul85 [17]
2 years ago
15

750.0g of water that was just boiled (heated to 100.0/c loses 78.45KJ of heat as it cools.What is the final temperature of water

?
Chemistry
1 answer:
Iteru [2.4K]2 years ago
4 0
  • Mass=m=750g=0.75kg
  • Q=78.45\times 10^3J=78450J
  • T_i=100°C
  • T_f=?
  • Specific heat capacity=c=4200J/kg°C

According to thermodynamics

\\ \tt\bull\rightarrowtail Q=mc\Delta T

\\ \tt\bull\rightarrowtail Q/mc=\Delta T

\\ \tt\bull\rightarrowtail \Delta T=\dfrac{78450}{0.75(4200)}

\\ \tt\bull\rightarrowtail \Delta T=24.9°C

\\ \tt\bull\rightarrowtail T_i-T_f=24.9

\\ \tt\bull\rightarrowtail T_f=100-24.9=76.1°C

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18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
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