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kkurt [141]
3 years ago
5

135-g sample of a metal requires 2.50 kJ to change its temperature from 19.5 C to 100.0 C what is the specific heat of this meta

l?
Chemistry
1 answer:
Harman [31]3 years ago
3 0
The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5. 
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Find ∆G for a cell whose potential is +0.24 V with 4 moles of electrons exchanged.
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(Yield Problem)
alex41 [277]

Answer:

Percent Yield Fe  =  82.5%

Explanation:

The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.

To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.

Atomic Mass (Mg): 24.305 g/mol

Atomic Mass (Fe): 55.845 g/mol

3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂

20.5 g Mg           1 mole              2 moles Fe            55.845 g
-----------------  x  -----------------  x  ----------------------  x  -----------------  =  
                           24.305 g           3 moles Mg             1 mole

=  31.4 g Fe

                                     Actual Yield
Percent Yield  =  ----------------------------------  x  100%
                                 Theoretical Yield

                               25.9 g Fe
Percent Yield  =  --------------------  x  100%
                               31.4 g Fe

Percent Yield  =  82.5%

5 0
2 years ago
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