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Andreas93 [3]
3 years ago
12

a piece of metal is heated to 100℃ and placed into 75.0g of water in a coffee-cup calorimeter. Initially, the temperature of the

water in the calorimeter was 23.1℃. The final temperature of he system was 30.6℃. What is the mass of the piece of metal? The specific heat of the water is 4.184 J/g℃. The specific heat of the metal is 0.900J/g℃.
Chemistry
1 answer:
Jet001 [13]3 years ago
4 0

Answer:

The mass of this piece of metal is 37.68g

Explanation:

<u>Step 1:</u> Given data

q = m*ΔT *Cp

⇒with m = mass of the substance

⇒with ΔT = change in temp = final temperature T2 - initial temperature T1

⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = 0.9J/ g °C)

<u>Step2 : </u>

For this situation : we get for q = m*ΔT *Cp

q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)

-m (30.6 -100) * 0.9 = 75g *( 30.6 - 23.1) * 4.184J / g °C

-m * (-69.4) * 0.9 = 75g * (7.5) * 4.184J / g °C

-m * (-69.4) *0.9 = 2353.5

-m = 2353.5 / (-69.4 *0.9) = -37.68

m = 37.68g

The mass of this piece of metal is 37.68g

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Explanation:

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(b) Calculate the amount of oxygen given by 0.361 g of KClO_{3} as follows.

    \frac {96g}{245.2g} \times 0.361 g = 0.141 g of O_{2}


c) Calculate the amount of oxygen given by 83.6 kg KClO_{3} as follows.  

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Convert kg into grams as follows.

    \frac{32.731 kg}{1 kg} \times 1000 g = 32731 g of O_{2}


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    \frac {96g}{245.2g} \times 22.5 mg = 8.79 mg

Convert mg into grams as follows.

  \frac{8.79 mg}{1 mg}\times 10^{-3} g = 8.79 \times 10^{-3} g of O_{2}

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