There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
it is equal to the potential energy of the products
Answer:
by heart beat camel lives
86 percent is the percent yield for this experiment if he expected to produce 5g of product.
Explanation:
Given that:
mass of test tube = 5 grams
mass of test tube + reactant is 12.5 grams
mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)
mass of reactant = 12.5 -5
= 7.5 grams
when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.
so mass of product formed = 9.3 - 5
= 4. 3 grams of product is formed (actual yield)
However, he expected the product to be 5 grams (theoretical yield)
Percent yield =
x 100
putting the values in the formula:
percent yield =
x 100
= 86 %
86 percent is the percent yield.