Ernest Rutherford's gold foil experiment was when he shot positively charged particles at a thin layer of gold foil. He concluded that there was a core that contained most of the atom's mass, the core is the nucleus
<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>
In stars helium atoms are used with hydrogen atoms in order to produce energy.
Answer: A )
I think it’s C but I’m not sure
