Answer:
a) The work done is 10.0777 kJ
b) The water's change in internal energy is -122.1973 kJ
Explanation:
Given data:
1 mol of liquid water
T₁ = temperature = 100.9°C
P = pressure = 1 atm
Endothermic reaction
T₂ = temperature = 100°C
1 mol of water vapor
VL = volume of liquid water = 18.8 mL = 0.0188 L
VG = volume of water vapor = 30.62 L
3.25 moles of liquid water vaporizes
Q = heat added to the system = -40.7 kJ
Questions: a) Calculate the work done on or by the system, W = ?
b) Calculate the water's change in internal energy, ΔU = ?
Heat for 3.25 moles:

The work done:

The change in internal energy:

Answer:
17
Explanation:
the atomic mass is away protons + neutrons, electrons are neglatable.
Answer:
In the Lewis structure of P4 there are 6 bonding pairs and 4 lone pairs of electrons.
Explanation:
The structure of tetrahedral molecule of P4 is provided below.
Each phosphorus atom has 5 valence electrons out of which 3 electrons involve in bonding and the rest 2 electrons exist as a lone pair that does not involve in bonding.Hence each phosphorus atom has one lone pair.In P4 molecule there are phosphorus atoms and hence 4 lone pairs in total.
As you can see in the figure, each phosphorus atom is bonded to the other three atoms.A bond is formed when two atoms share one electron each and the pair is called bonding pair.
From the figure we can see that there are 6 bonds in total.Each bond consist of one bonding pair of electrons and hence in total there are 6 bonding pairs of electrons.
Hence in a P4 molecule there are six bonding pairs and 4 lone pairs of electrons.
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate