Question 1: A 42.1g peice of metal was heated up to 97.4C and then dropped into

Question

Question 1: A 42.1g peice of metal was heated up to 97.4C and then dropped into a beaker containing 63.0g of water at 22.0C. When the water and metal come to thermal equilibrium, the temperature is 30.10C. What is the specific heat capacity of the metal? Know the specfic heat capcity of the water is 4.184 J/(g-K)?

Multiple Choice:

A) 0.980 J/(g-K)

B) 0.600 J/(g-K)

C) 0.488 J/(g-K)

D) 0.387 J/(g-K)

E) 0.754 J/(g-K)

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Question 2: In which of the following is there a violation of normal oxidation number rules?

A) BaO2

B) RaO2

C) H2O2

D) None of these.

E) All of them.

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Explanation / Answer

They become equal at 30.10
I'm assuming you're familar with M.C./T
Simply use M.C./T = M.C./T

0.0421 x C x (97.4--30.1) = 0.063 x 4.184 x (30.1 - 22)

2.833*c=2.14

C=0.754

option=E

BaO2; Ba +2 ,O -1

H-O-O-H
The net formal charge on each of oxygen atom will be -1(when electrons are distributed as per convention.

So oxidation number in this case for bothe the oxygen atoms will be -1.

RaO2; Ra +2, O -1

OPTION-D

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