<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
Answer:
0.186M
Explanation:
First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:
<em>Moles Nitric acid:</em>
0.0200L * (0.100mol / L) = 0.00200 moles
0.120L * (0.200mol / L)= 0.02400 moles
Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid
Molarity: 0.026 moles / 0.140L
<h3>0.186M</h3>
They are DIFFUSION and OSMOSIS.
Answer:
hexaaquairon(III) trinitrate
