Calculate the heat (in calories) required to freeze 35.0 g of water.

-2.79 × 10³ cal

Explanation:

Step 1: Given data

Step 2: Calculate the heat required to freeze 35.0 g of water

We have 35.0 g of liquid water and we want to freeze it, that is, to convert it in 35.0 g of ice (solid water), at 0 °C (melting point). We can calculate the heat (Q) that must be released using the following expression.

Q = L × m

Q = -79.7 cal/g × 35.0 g

Q = -2.79 × 10³ cal

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6