There are exactly (a). 10.0 and (b). 28.0
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Answer:

Explanation:
= Mass of metal = 19 g
= Specific heat of the metal
= Temperature difference of the metal = 
V = Volume of water = 150 mL = 
= Density of water = 
= Specific heat of the water = 4.186 J/g°C
= Temperature difference of the water = 
Mass of water

Heat lost will be equal to the heat gained so we get

The specific heat of the metal is
.
I cant entirely tell for now but an article on rodioactivity should solve the problem
Answer: I think it represents the SiO
Explanation: