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adell [148]
3 years ago
5

4.07 x 10-17 (5.6 x 10") (5.8 x 105)

Chemistry
1 answer:
Andrej [43]3 years ago
3 0
4.07 x 10-17 = 23.7
(5.6 x 10”) (5.8 x 105)= log(100)
log(7)
Hope this helped
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VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
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Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

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m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

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T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

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The answears are in the attached photo.

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