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myrzilka [38]
3 years ago
7

A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r

ises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J/ (°C × g), what is the specific heat of the metal?
Chemistry
1 answer:
vampirchik [111]3 years ago
5 0
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)
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Explanation:

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What is the theoretical yield of NaBr
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The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

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Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

Learn more about stoichiometry:

brainly.com/question/14735801

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