A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final temperature of the water is 31.5 oC. What is the heat change of the metal in joules? The specific heat of water is 4.184 J/goC

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

Heat change of the metal = -6799 J

Explanation:

Step 1: Data given

Mass of the metal = 20.0 grams

Mass of water = 250.0 grams

Initial temperature of water = 25.0 °C

Final temperature of water = 31.5 °C

Step 2: Calculate the heat change of the metal in joules?

Heat lost =  heat gained

Qlost = -Qgained

Q = m*C*ΔT

m(metal) * C(metal) * ΔT(metal) = -m(water) * C(water) * ΔT(water)

⇒with m(metal) = the mass of metal = 20.0 grams

⇒with C(metal) = the specific heat of meteal = ?

⇒with ΔT(metal) = the change of temperature of the metal = ?

⇒with m(water) = the mass of water = 250.0 grams

⇒with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 31.5 °C - 25.0 °C = 6.5 °C

The heat change of metal = heat change of water

Q = 250 * 4.184 * 6.5

Q = 6799 J

Heat change of water = 6799 J

Heat change of the metal = -6799 J

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

indium, and thallium have three electrons i think