Answer:
can you show a clearer picture of the chart?
Explanation:
1 mole of h3bo3...........6.023*10²³ each h and B and 0 so we will have 3hydrogen+ 1 B+3 oxygen = 7*6.023*10²³ atoms
1 mole .......7*6.023*10²³atoms
4 moles ........x atoms
x=4*7*6.023*10²³.
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Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample