I’m really sorry I need points I hope you find an answer
<span>83.9%
First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements.
Atomic weight aluminum = 26.981539
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol
Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol
Now determine how many moles of C6H6 was produced
Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol
Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required.
0.012174851 mol / 3 = 0.004058284 mol
Then multiply by the molar mass to get the number of grams that was originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So
1.048227218 g / 1.25 g = 0.838581775 = 83.8581775%
And of course, round to 3 significant digits, giving 83.9%</span>
I would say copper, silver, and tin, since an alloy is a mixture of metals and metalloids.
Answer: The number of N atoms in 137.0 g of N2O3 21.67 x 10∧23 atoms.
Explanation:
- We must obtain the number of moles of the compound: (n = mass/molar mass), mass = 137.0 g and molar mass of N2O3 = 76.01 g/mol.
- n = (137.0 g)/ (76.01 g/mol) = 1.80 mol.
- It is necessary to determine the number of molecules of this sample.
- Every mole contains Avagadro's number (6.02 x 10^23) of molecules.
- The number of molecules = (6.02 x 10^23)(1.80) = 10.84 x 10∧23 molecules.
- Every molecule of N2O3 contain 2 atoms of N.
- The number of N atoms in 137.0 g of N2O3 = (10.84 x 10∧23 molecule) (2 atoms) = 21.67 x 10∧23 atoms.
