just divide the mass 49.2 grams by the amount of gummi bears 23 to get the answer of 2.1391 then it says to round to 2 decimals so 2.14 grams
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
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First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
The answer to your question is full