Near water, change in elevation, or change in latitude.
<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>
After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Hey there!
Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance PO₄.
1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂
Balance H.
6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.
Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Balance Ca.
1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Our final balanced equation:
3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂
Hope this helps!
