<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
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Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
Answer: Aluminum, Antimony, Arsenic, Bismuth, Carbon, Cadmium, Chromium, Cobalt, etc.
There are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample. Details about number of molecules can be found below.
<h3>How to calculate number of molecules?</h3>
The number of molecules of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.
According to this question, there are 2000g of ZnO in a sample. Zinc oxide has a molar mass of 81.38 g/mol.
no of moles = 2000g ÷ 81.38g/mol
no of moles = 24.57mol
number of molecules = 24.57 × 6.02 × 10²³
number of molecules = 147.95 × 10²³
Therefore, there are 1.48 × 10²⁵ molecules of zinc oxide in a 2 kg sample.
Learn more about number of molecules at: brainly.com/question/11815186
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Temperature Decreases
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