the balanced equation for the formation of ammonia is
N₂ + 3H₂ ---> 2NH₃
molar ratio of N₂ to NH₃ is 1:2
mass of N₂ reacted is 8.0 g
therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol
according to the molar ratio,
1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant
therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃
therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g
a mass of 9.72 mol of NH₃ is formed
Answer:
24.309 g/mol
Explanation:
To get the atomic mass, all we have to do is calculate with the masses of the three isotope, the real quantity present, taking account of the percent and then, do a sum of these three values. Like a pondered media.
For the first isotope:
23.99 * (78.99/100) = 18.95 g/mol
For the second isotope:
24.99 * (10/100) = 2.499 g/mol
For the last isotope:
25.98 * (11.01/100) = 2.86 g/mol
Now, let's sum all three together
AW = 18.95 + 2.499 + 2.86
AW = 24.309 g/mol
E.g. in H3PO4 (O, -2).
8. The sum of the oxidation states of all the atoms in a species must be equal to net charge on the species. e.g. Net Charge of HClO4 = 0, i.e. [+1(H)+7(Cl)-2<span>*4(O)] = 0.</span>
Don't take my word for it but I think it is
1: proteins
2: energy from the sun, carbon dioxide, and water
3: this one is confusing me but I think it would be nutrients from food and oxygen
4: water
Again these are attempts I can't prove these
due to there reactive rate?
