Q.34 The temperature of a 12.58 g sample of calcium carbonate increases from 23.6 °C to 38.2C. If the specific heat of calcium carbonate is 0.82 J/g-K, how many joules of heat are absorbed?

150.6 Joules

Explanation:

Calorimetry problem must be solved by the calorimetry formula.

Q = c . m . ΔT

ΔT = Final T° - Initial T°

c = specific heat (Notice, that in the unit we have K as unit for T°, and the T° is in °C. We don't need to make the conversion, because is a Δ)

For example 100°C - 50°C = 50°C

373K - 323K = 50 K

So let's replace data given: Q = 0.82 J/g.K . 12.58 g . (38.2 - 23.6)

Q = 150.6 Joules

We assume no heat lost by sourrounding

i think a, hope it .

explanation:

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms