16.94/18=.9411111
sig figs: 0.9411 mole of water
Mass percent= grams solute/ grams of solution x 100
Mass Percent= (50/ 150)x100= 33.3%
Answer:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen = 
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,
solving for x, we get:
<u>x = 324.18 g/mol</u>
Metal rusts when it oxidized around moisture.
Answer:
Percent yield = 90.5%
Explanation:
Given data:
Mass of carbon dioxide = 500 g
Mass of water = excess
Actual yield of carbonic acid = 640 g
Percent yield = ?
Solution:
Balanced chemical equation:
CO₂ + H₂O → H₂CO₃
Number of moles of carbon dioxide
Number of moles = Mass / molar mass
Number of moles = 500 g/ 44 g/mol
Number of moles = 11.4 mol
Now we will compare the moles of H₂CO₃ with CO₂.
CO₂ : H₂CO₃
1 : 1
11.4 : 11.4
Mass of carbonic acid:
Mass = number of moles × molar mass
Mass = 11.4 mol × 62.03 g/mol
Mass = 707.14 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 640 g/ 707.14 g × 100
Percent yield = 90.5%
