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Annette [7]
3 years ago
8

A 1.25g sample of copper (cCu=0.386Jg∘C) is initially at a temperature of 25.0∘C. If the sample absorbs 87.4J of heat, what is i

ts final temperature? Report your answer with the correct number of significant figures.
Chemistry
1 answer:
professor190 [17]3 years ago
4 0

Answer: Final temperature = 206∘C

Explanation:

Heat Energy is given as  

q= mCΔT

ehere

q= Heat energy = 87.4J

m= mass=1.25g

C=specific heat c= 0.386Jg∘C) ,

ΔT =  Change in temperate of which the final temperature= 25.0∘C

 q= mCΔT

ΔT = q/mC

ΔT = 87.4/ 1.25 X 0.386=181.14∘C

But,

T final- T initial = ΔT

T final = T initial + ΔT

T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C

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- 13.56 g of sodium chloride are theoretically yielded.

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Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

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Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

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