It simply equals 4x-16.......
the expression is in lowest terms, so you can’t do anything else
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
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
Explanation:
(A)role of nittogen fixing bacteria
=Nitrogen-fixing bacteria, microorganisms capable of transforming atmospheric nitrogen into fixed nitrogen (inorganic compounds usable by plants). More than 90 percent of all nitrogen fixation is effected by these organisms, which thus play an important role in the nitrogen cycle.
B)role of nitrifying bacteria
=Nitrifying bacteria convert the most reduced form of soil nitrogen, ammonia, into its most oxidized form, nitrate. In itself, this is important for soil ecosystem function, in controlling losses of soil nitrogen through leaching and denitrification of nitrate.
C)role of denitrifying bacteria
=Denitrifying bacteria converts nitrates back to nitrogen gas.
Answer: 2. they are 4 and 8
Explanation: I took the exam. I hope I helped :)
