Answer:
0.09 × 10²³ molecules of O₂
Explanation:
Given data:
Mass of mercury(II) oxide = 6.54 g
Molecules of oxygen produced = ?
Solution:
Chemical equation:
2HgO → 2Hg + O₂
Number of moles of HgO:
Number of moles = mass/ molar mass
Number of moles = 6.54 g / 216.6 g/mol
Number of moles = 0.03 mol
Now we will compare the moles of HgO with oxygen.
HgO : O₂
2 : 1
0.03 : 1/2×0.03 = 0.015 mol
Number of molecules of oxygen:
1 mole of O₂ = 6.022 × 10²³ molecules of O₂
0.015 mole of O₂ :
0.015 × 6.022 × 10²³ molecules of O₂
0.09 × 10²³ molecules of O₂
The rule that we will use to solve this problem is:
M1V1 + M2V2 + M3V3 = MtVt
where:
M1 is the molarity of first solution = 0.003 molar
V1 is the volume of the first solution which is the X that we want to find
M2 is the molarity of second solution = 0.0001 molar
V2 is the volume of second solution = 20 ml
M3 is the molarity of third solution = 0.002 molar
V3 is the volume of the third solution = 30 ml
Mt is the final total molarity = 0.0015 molar
Vt is the final volume = V1+V2+V3 = X+20+30 = (X+50) ml
Substitute with the above givens in the equation to get X as follows:
0.003X + 0.0001(20) + 0.002(30) = (X+50)(0.0015)
0.003X + 0.062 = 0.0015X + 0.075
0.003X - 0.0015X = 0.075 - 0.062
0.0015X = 0.013
X = 0.013 / 0.0015
X = 8.667 ml
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Answer:
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