Answer:
Explanation:
<u>Molecular formula from Glucose:</u>
C₆H₁₂O₆
<u>3 moles of Glucose:</u>
3C₆H₁₂O₆
In 1 mole of Glucose, there are 12 hydrogen atoms.
<u>In 3 moles:</u>
= 12 × 3
= 36 H atoms
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Answer #1 is "there is 2.5 grams of solute in every 100 g of solution."
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution."
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
To answer this item, we assume that oxygen behaves ideally such that it is able to fulfill the following equation,
PV = nRT
If we are to retain constant the variable n and V.
The percent yield can therefore be solved through the following calculation,
n = (10.5 L)/(22.4 L) x 100%
Simplifying,
n = 46.875%
Answer: 48.87%
