Answer:
Explanation:
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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
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There are 4 significant figures! Start counting after the first non-zero digit :)
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Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
Answer:
Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.
Explanation:
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