<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>
(B), because 1.0 moles would be 6.02 x 10^23 molecules. So you have half a mole.<span>
</span>
Answer:
6.022 x 10²³
Explanation:
The number of atoms = the number of moles x with the Avogadro's number.
(The Avogadro's number is 6.022 x 10²³ atoms / moles)
number of atoms = 1.00 moles x 6.022 x 10²³ atoms / mole
number of atoms = 6.022 x 10²³ atoms
(There is no need to simplify?) = 6.022 x 10²³
(ps. This is my first time doing this question so im sorry if i got it wrong
(つ﹏⊂)
Answer:
0.978 M
Explanation:
Given data
- Mass of luminol (solute): 13.0 g
- Volume of the solution = volume of water: 75.0 mL = 0.0750 L
We can find the molarity of the stock solution of luminol using the following expression.
M = mass of solute / molar mass of solute × liters of solution
M = 13.0 g / 177.16 g/mol × 0.0750 L
M = 0.978 M
