Answer:
The empirical formula of ribose (a sugar) is CH2O. In a separate experiment, using a mass spectrometer, the molar mass of ribose was determined to be 150 g/mol.Explanation:
Answer:
<u>5 moles S x (36.02 g S/mole S) = 180.1 grams of S</u>
Explanation:
The periodic table has mass units for every element that can be correlated with the number of atoms of that element. The relationship is known as Avogadro's Number. This number, 6.02x
, is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x
. <u>The mole is correlated to the atomic mass of that element.</u> The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. <u>But it can also be read as 36.02 grams/mole.</u>
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<u>This means that 36.02 grams of S contains 1 mole (6.02x</u>
<u>) of S atoms</u>.
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This relationship holds for all the elements. Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole. ^5.38 grams of Zn contains 1 mole of Zn atoms.
And so on.
5.0 moles of Sulfur would therefore contain:
(5.0 moles S)*(36.02 grams/mole S) = <u>180.1 grams of S</u>
Note how the units cancel to leaves just grams. The units are extremely helpful in mole calculations to insure the correct mathematical operation is done. To find the number of moles in 70 g of S, for example, we would write:
(70g S)/(36.02 grams S/mole S) = 1.94 moles of S. [<u>Note how the units cancel to leave just moles</u>]
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
Answer:
97.78% KCl in the original sample
Explanation: