Answer:
a) Change in temperature of the freezing point is +5.35C.
b) The molality of the hydrocarbon is 0.265 M
c) Mass of cyclohexane is 0.014354 g
d) The number of moles of hydrocarbon in the solution is
e) Molar mas of hydrocarbon is .
Explanation:
a)
Therefore, Change in temperature of the freezing point is +5.35C.
b)
Therefore, The molality of the hydrocarbon is 0.265 M.
c)
Therefore, Mass of cyclohexane is 0.014354 g
d)
Therefore, The number of moles of hydrocarbon in the solution is
e)
Therefore, Molar mas of hydrocarbon is .
Answer:
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Explanation:
Answer:
2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂
Explanation:
The balanced reaction is:
Pb(CH₃COO)₂ + H₂S → 2 CH₃COOH + PbS
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) they react and produce:
- Pb(CH₃COO)₂: 1 mole
- H₂S: 1 mole
- CH₃COOH: 2 moles
- PbS: 1 mole
In this case, to know how many grams of H₂S are needed to produce 18.00 g of PbS, it is first necessary to know the molar mass of the compounds H₂S and PbS and then to know how much it reacts by stoichiometry. Being:
- H: 1 g/mole
- S: 32 g/mole
- Pb: 207 g/mole
The molar mass of the compounds are:
- H₂S: 2* 1 g/mole + 32 g/mole= 34 g/mole
- PbS: 207 g/mole + 32 g/mole= 239 g/mole
So, by stoichiometry they react and are produced:
- H₂S: 1 mole* 34 g/mole= 34 g
- PbS: 1 mole* 239 g/mole= 239 g
Then the following rule of three can be applied: if 239 grams of PbS are produced by stoichiometry from 34 grams of H₂S, 18 grams of PbS from how much mass of H₂S is produced?
mass of H₂S= 2.56 grams
<u><em>2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂</em></u>
B is the right one. They aren't compounds, which is a mix of two elements.
Neutral pH means a pH of 7, yes? Recall that 'pH' means the POWER of Hydrogen. You should be familiar with the equation : pH = -log[H+].
In this case, 7 = log[H+].
Mathematically rearranged, this means that to find H+ concentration,
[H+] = 10 ^-pH.
In this case, [H+] = 10 ^-7.
I am assuming this is conducted at a standard 25 degrees. The auto-dissociation formula of water is this: Kw = [H3O+] x [OH-] at 25 degrees.
The concentration of Hydronium is the same as the concentration of hydrogen, 10 ^-7. mol/L.
There's your answer!