Answer:
0.440 moles of NH₃ are produced
Explanation:
First of all, we need to determine the limiting reactant by the stoichiometry.
Equation reaction is: N₂(g) + 3H₂(g) ⟶ 2 NH₃(g)
1 mol of nitrogen needs 3 moles of hydrogen to react
Therefore 0.220 moles of N₂ will need (0.220 . 3) / 1 = 0.660 moles of H₂
As we have 0.717 moles of H₂ and we need 0.660, the hydrogen is the excess reagent, therefore, the N₂ is the limiting reactant
3 moles of H₂ need 1 mol of N₂ to react
Then, 0.717 moles of H₂ will react with (0.717 . 1) / 3 = 0.239 moles of N₂
We do not have enough N₂
After complete reaction → ratio is 1:2
1 mol of N₂ reacts to produce 2 moles of ammonia
Therefore 0.220 moles of N₂ will produce (0.220 . 2) / 1 = 0.440 moles of NH₃
That is false. Weight is not constant every place in the universe. Take space for example. It all depends on gravity. So, the answer is false.
Answer:
81.71%
Explanation:
One mole of propane contains 3 moles of carbon atoms and 8 moles of hydrogen atoms, as seen from the molecular formula of 
. In order to calculate the percent of carbon in propane by mass, we need to remember that %w/w (or percent mass) formula states that:
That is, we need to divide the mass of the component of interest by the total mass of the compound and multiply by 100 to obtain the percentage.
For simplicity, let's take 1 mole of propane and find the mass of 1 mole (hence, we'll be finding the molar mass of propane). To do that, we add the 3 molar masses of carbon and 8 molar masses of hydrogen to obtain a total of:
Now that we have the molar mass of propane, we also need to find the total mass of carbon in 1 mole of propane. We know that we have a total of 3 moles of carbon which corresponds to:
Dividing the mass of carbon present by the total mass of the compound will yield the mass percentage as defined by the formula we introduced:
Answer:
while analogous structures show that similar selective pressures can produce similar adaptations (beneficial features). Similarities and differences among biological molecules can be used to determine species' relatedness.
Explanation:
