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2 years ago

CuFeS2 + 3 CuCl2 --> 4 CuCl + FeCl2 + 2 S How many mL of 0.075 M CuCl2 are needed to titrate 0.20 grams of CuFeS2?

Chemistry

1 answer:

GuDViN [60]2 years ago

5 0

Answer:

$V_{CuCl_2 }= 43.6mL$

Explanation:

Hello there!

In this case, according to this titration problem, it is possible to note there is a 1:3 mole ratio of CuFeS2 to CuCl2; it means that we can use the following equation:

$3*n_{CuFeS_2}=n_{CuCl_2 }$

Thus, by introducing the mass and molar mass of the former and the volume and concentration of the latter, we write:

$3*0.20gCuFeS_2*\frac{1mol}{183.51g} =V_{CuCl_2 }*0.075\frac{mol}{L} \\\\V_{CuCl_2 }=\frac{0.00327mol}{0.075mol/L}*\frac{1000mL}{1L} \\\\V_{CuCl_2 }= 43.6mL$

Regards!

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6. The graph below shows the heating curve for ethanol (from –200C to 150C). Calculate the amount of heat (kJ) required for each

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This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

$Q_T=Q_1+Q_2+Q_3+Q_4+Q_5$

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

$Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\
\\
Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\
\\
Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\
\\
Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\
\\
Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ$

Finally, we add them up to get the result:

$Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\
\\
Q_T=1.758kJ$

Learn more about heating curves: brainly.com/question/10481356

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