How many kilojoules of heat are absorbed when 2.00 L of water is heated from 19*C to 80*C? (Show all work including the equation you use to solve)

The answer to your question is 510.7 kJ

Explanation:

Data

Q = ?

Volume = 2 L

Temperature 1 = T1 = 19°C

Temperature 2 = T2 = 80°C

Specific heat = 4.186 J/kg°C

Density of water = 1 g/ml

Equation

              Q = mC(T2 - T1)

Process

1.- Calculate the mass of water

mass = density x volume

mass = 1 x 2000

mass = 2000 g

2.- Substitute values in the Heat formula

              Q = (2000)(4.186)(80 - 19)

-Simplification

               Q = (2000)(4.186)(61)

-Result

              Q = 510692 J  or 510.7 kJ

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.