Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16
Hope this helps
Answer:
1.6 grams
Explanation:
We need to prepare 100 mL (0.100 L) of a 0.10 M CuSO₄ solution. The required moles of CuSO₄ are:
0.100 L × 0.10 mol/L = 0.010 mol
The molar mass of CuSO₄ is 159.61 g/mol. The mass corresponding to 0.010 moles is:
0.010 mol × (159.61 g/mol) = 1.6 g
We should use 1.6 grams of CuSO₄.
<span>At 100 feet, the diver is under about 4 atmospheres pressure. If she is free diving, her lungs will be compressed to about 1/4 their size on the surface (with some movement of the major abdominal organs). If she is scuba diving, the air which she is breathing is also at 4 atmospheres and there is no problem. (The non-gas spaces in the body are not-compressible and are unaffected.) The only problems she has to concern herself with are the beginnings to nitrogen narcosis and the nitrogen which is dissolving (Henry's law) into her body tissues. On the way up, she also has to remember that the air in her lungs will expand by a factor of 4 and she better exhale! Hope this helps you</span>
The answer is A. you sre correct!
<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>
