Answer: The first steps
Explanation: science
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 ![3.8\times 10^{-3}moles](https://tex.z-dn.net/?f=3.8%5Ctimes%2010%5E%7B-3%7Dmoles)
e) Molar mas of hydrocarbon is
.
Explanation:
a)
![\Delta T = Freezing\,point\,of\,hydrocarbon+Freezing\,point\,of\,pure\,cyclohexane](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20Freezing%5C%2Cpoint%5C%2Cof%5C%2Chydrocarbon%2BFreezing%5C%2Cpoint%5C%2Cof%5C%2Cpure%5C%2Ccyclohexane)
![\Delta T = -1.25+6.60=+5.35^{o}C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20-1.25%2B6.60%3D%2B5.35%5E%7Bo%7DC)
Therefore, Change in temperature of the freezing point is +5.35C.
b)
![\Delta T= K_{F}\times Molality](https://tex.z-dn.net/?f=%5CDelta%20T%3D%20K_%7BF%7D%5Ctimes%20Molality)
![Molality=\frac{5.35^{o}C}{20.2^{o}C}=0.265m](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7B5.35%5E%7Bo%7DC%7D%7B20.2%5E%7Bo%7DC%7D%3D0.265m)
Therefore, The molality of the hydrocarbon is 0.265 M.
c)
![Mass\,of\,cyclohexane=171.206-156.852=14.354gm=0.01435Kg](https://tex.z-dn.net/?f=Mass%5C%2Cof%5C%2Ccyclohexane%3D171.206-156.852%3D14.354gm%3D0.01435Kg)
Therefore, Mass of cyclohexane is 0.014354 g
d)
![Molality=\frac{Number\,of\,moles}{Mass\,of\,cyclohexane}](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7BNumber%5C%2Cof%5C%2Cmoles%7D%7BMass%5C%2Cof%5C%2Ccyclohexane%7D)
![Number\,of\,moles=0.265\times 0.014354=3.8\times 10^{-3}moles](https://tex.z-dn.net/?f=Number%5C%2Cof%5C%2Cmoles%3D0.265%5Ctimes%200.014354%3D3.8%5Ctimes%2010%5E%7B-3%7Dmoles)
Therefore, The number of moles of hydrocarbon in the solution is ![3.8\times 10^{-3}moles](https://tex.z-dn.net/?f=3.8%5Ctimes%2010%5E%7B-3%7Dmoles)
e)
![Molar\,mass=\frac{0.300g}{3.8\times 10^{3}mole}=78.9g.mol^{-1}](https://tex.z-dn.net/?f=Molar%5C%2Cmass%3D%5Cfrac%7B0.300g%7D%7B3.8%5Ctimes%2010%5E%7B3%7Dmole%7D%3D78.9g.mol%5E%7B-1%7D)
Therefore, Molar mas of hydrocarbon is
.
Answer:
A. Mg(OH)2
Explanation:
all oxides are insoluble except those of Group IA metals. most sulfides S2- are insoluble, with the exceptions of Group I, II (slightly soluble) metals and ammonium - NH4+
Answer:
pKa of the acid is 3.6
Explanation:
When a weak acid, HX, reacts with NaOH, the conjugate base, X⁻, is produced:
HX + NaOH → X⁻ + Na⁺ + H₂O
At the half neutralized solution, [HX] = [X-]
Based on Henderson-Hasselbalch equation:
pH = pKa + log [ X⁻] / [HX]
<em>Where pH is the pH of the buffer = 3.6</em>
<em>pKa is the pka of the solution</em>
<em>And as [ X⁻] = [HX], [ X⁻] / [HX] = 1</em>
<em />
Replacing:
3.6 = pKa + log 1
3.6 = pKa + 0
<h3>pKa of the acid is 3.6</h3>
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