A molecule that can h-bond will not always necessarily and does not have guarantee to have a higher boiling point than one than cannot have h-bond.
we can take an example of Pentan-2-one that cannot h-bond but instead of this it has a high boiling point that is 102.3 °C, while propan-1-ol can h-bond but it has a boiling point of 97.2°C, that is lower than the boiling point of Pentan-2-one.
140 g of nitrogen (N₂)
Explanation:
We have the following chemical equation:
N₂ + 3 H₂ -- > 2 NH₃
Now, to find the number of moles of ammonia we use the Avogadro's number:
if 1 mole of ammonia contains 6.022 × 10²³ molecules
then X moles of ammonia contains 6.022 × 10²⁴ molecules
X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³
X = 10 moles of ammonia
Taking in account the chemical reaction we devise the following reasoning:
If 1 mole of nitrogen produces 2 moles of ammonia
then Y moles of nitrogen produces 10 moles of ammonia
Y = (1 × 10) / 2
Y = 5 moles of nitrogen
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of nitrogen (N₂) = 5 × 28 = 140 g
Learn more about:
Avogadro's number
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Answer:
16.9g
Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2
Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.
Then to calculate the theoretical yield you need to compare moles of the reactants:
m(Cu)=5g
M(Cu)=63.55
n(Cu)=5/63.55=0.0787
By comparing coefficients you require twice as much silver: 0.157mol
n(Ag)=0.157
M(Ag)=107.86
m(Ag)=0.157x107.86=16.9g
Hence, the theoretical yield of this reaction would be 16.9g
