Part A How much time (in minutes) will it take for the water heater to heat the 5.5 L of water from 23 ∘C to 43 ∘C? (Assume that the water has a density of 1.0 g/mL.)

Time = t = 30.01 minutes

Explanation:

The missing part of the question is that "The water heater power rating is 255 Watts, where 1 Watt = 1 J/s.

Firstly, calculating the work done by water heater to heat 5.5 L of water from 23° C to 43°C.

Work Done = Q = m C ΔT

where,

Q = Heat Transfer

m = mass of water

m = Density x Volume = 1 g/mL x 5.5 L = 1000g/L x 5.5L = 5500 g

ΔT = Temperature Difference = 43-23 = 20°C

C = Heat Capacity of Water = 4.186 J/g °C

Therefore,

Since, Power can be given as Work done per unit time,

or, in this case,

Here,

P = 255W or P = 255J (Since 1W = 1J)

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

the answer is b & c i believe : )