Answer:
N2 + 3H2 ———> 2NH3
As we know 1000 grams ammonia is 58.82 moles so according to unitary method,
2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.
No. Of moles = given mass/molar mass
Implies that
Mass of nitrogen required = 29.41*28 = 823.48 grams.
Explanation:
Answer:
Noble Gases
Explanation:
The Noble Gases have a full valence shell of 8 electrons. They are stable in that sense.
<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
High tide or Neap tides which are normally really high.