D. CuCl2 copper(2)chloride
Answer:
1. Nuts
2. Canned meats and seafood
3. Dried grains
4. Dark chocolate
5. Protein powders
Answer:
The heat capacity for the sample is 0.913 J/°C
Explanation:
This is the formula for heat capacity that help us to solve this:
Q / (Final T° - Initial T°) = c . m
where m is mass and c, the specific heat of the substance
27.4 J / (80°C - 50°C) = c . 6.2 g
[27.4 J / (80°C - 50°C)] / 6.2 g = c
27.4 J / 30°C . 1/6.2g = c
0.147 J/g°C = c
Therefore, the heat capacity is 0.913 J/°C
