How many joules of heat are absorbed when the tempurature of a 13.9 g sample of caco3 (s) increases from 21.7c to 33.3 c? specific heat of calcium carbonate is .82 j/g-k

Q =132.22 j

Explanation:

Given data:

Mass of CaCO₃ = 13.9 g

Initial temperature = 21.7 °C

Final temperature = 33.3 °C

Specific heat of  CaCO₃ = 0.82 j/g.k

Amount of heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT =  33.3°C - 21.7°C

ΔT = 11.6°C

Q = m.c. ΔT

Q = 13.9 g. 0.82 j/g.k.  11.6°C

Q =132.22 j

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.