<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>
Answer:
6.0 L
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 0.075 M
V1 = 200 L
M2 = 2.5 M
V2 = ?
Solve for V2 --> V2 = M1V1/M2
V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L
Answer:
i belive the answer you are looking for is A.. hope this helps!
Explanation:
im not really 100% certain of this answer due to the fact i am a little rusty.
The answer is Heat energy moves from warmer objects to cooler ones
