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kodGreya [7K]
3 years ago
14

When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu

me that no heat is lost to the surroundings. What is the specific heat of the metal? The specific heat of water is 4.184 J/g·°C.
Chemistry
1 answer:
olga2289 [7]3 years ago
8 0

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC  = C = 0,50 J/gºC

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Answer:

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B. CuC₂O₄          Copper oxalate

C. Bi₂(C₂O₄)₃         Bismuth (III) oxalate

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This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

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It can be formed by the neutralization of the acid with the base

H₂C₂O₄  + 2KOH  → K₂C₂O₄  +  2H₂O

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This is a precipitate.

C.  2Bi³⁺  +  3C₂O₄⁻²   ⇄  Bi₂(C₂O₄)₃  ↓

This is a precipitate.

D.  Pb⁴⁺ +  2C₂O₄⁻²   ⇄  Pb(C₂O₄)₂  ↓

This is a precipitate.

E. 2NH₄⁺  +  C₂O₄⁻²   ⇄  (NH₄)₂C₂O₄  ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

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