After returning home from the grocery store, you mistakenly left a sealed mug of water (mass of H2O = 512g) on top of your car. When you got in your car the next day, the water was frozen and the mug had cracked. Answer the questions below using the following information: * melting point of water at 1.00 atm = 0.00°C * Cwater, liquid = 75.4 J/mol·°C

* DH°fus = 6.02 kJ/mol * Cwater, solid = 37.6 J/mol·°C

a. If the water in the mug was initially 25.5°C (approx. room temperature) and cooled to -30.0°C, draw a heating/cooling curve for this scenario (assume a pressure of 1.00 atm). Label your x- and y-axes, as well as the different sections of the curve (describe the physical state of water at these sections).

b. Calculate the energy change (in kJ) when the water cooled from 25.5°C to -30.0°C

answer: at room temperature (roughly 20 degrees c), you can dissolve 203.9 grams (or 200 grams) of sugar in 100 ml of water. as water temperature increases, the amount of sugar that will dissolve increases, up to about 500 grams of sugar in 100 ml water at 100 degrees c.

explanation:

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

false

because it does not have the potential to be entirely used up

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