Liquid ammonia (boiling point = –33.4C at 1 atm) can be used as a refrigerant and heat transfer fluid. How much energy is needed to heat 25.0 g of NH3(l) from –65.0C to –12.0C at 1 atm?

39.449 kJ.

Explanation:

q = m * Cp * Delta T

Where,

Delta T = temperature change in K

Cp = specific heat capacity in kJ/g.K

m = mass of the substance.

Delta T =33.4°C - (-65)°C

= 31.6 °C

q = 31.6 * 4.7 * 25

= 3713 J

= 3.713 kJ.

q = n * molar heat of vapourisation

Number of moles = mass/molar mass

= 25/17

= 1.47 mol.

q = 1.47 * 23500

= 34559 J

= 34.559 kJ.

q = m * Cp * Delta T

Delta T = -12°C - (-33.4)°C

= 21.4 °C

= 25 * 2.2 * 21.4

= 1177 J

= 1.177 kJ.

Adding all the energies together:

1.177 + 34.559 + 3.713

= 39.449 kJ

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.