A chemical process used to produce ethanol as a fuel additive was expected to produce 5,000 kilograms of ethanol based on the amounts of starting materials used, but only 4,760 kilograms were produced. What was the percent yield for ethanol in this process? A)1.09 percent B)4.80 percent C)95.2 percent D)105 percent

The percent yield for ethanol in this process is 95.2%

Explanation:

Yield is the ratio between the amount of product actually obtained in the reaction and the maximum amount of product that could have been obtained if the reagents had been completely consumed.

The theoretical yield is the yield is the maximum you can get in a reaction.

On the other hand, the actual performance shows what you really get in experimentation. It is generally less than the theoretical yield.

So, the percentage return is the actual return times the theoretical return multiplied by 100.

So, in this case the percentage yield will be calculated as:

percent yield=  95.2%

The percent yield for ethanol in this process is 95.2%

it's c. elements are placed in rows based on how may valencies electrons have.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.