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Naddik [55]
3 years ago
5

In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i

nto a cup containing 82.9 grams of water at 24.16°C. She measures the final temperature to be 25.70°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of zinc to be J/g°C.
Chemistry
1 answer:
fgiga [73]3 years ago
7 0

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

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Answer:

Explanation:

mole of NaOH present = molarity x volume

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From the equation of reaction:

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The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>

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