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Since Group 2 alkali earth metals have 2 valence electrons, they tend to lose those 2 when forming ionic bonds. And the Loss of Electrons = Oxidation (L.E.O. for short). Therefore this group, including Mg and Ca, have an oxidation of [+2].
So the correct answer is C) +2
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
The correct answer is D.
Increasing the pressure shifts the position of equilibrium towards the side with fewer gas molecules.
Answer:
C. 1.3 mol
Explanation:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
Given:
P = 121.59 kPa
V = 31 L
T = 360 K
R = 8.3145 L kPa / mol / K
Find: n
n = PV / (RT)
n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)
n = (3769.29 L kPa) / (2993.22 L kPa / mol)
n = 1.26 mol
Round to two significant figures, there are 1.3 moles of gas.
