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trapecia [35]
2 years ago
13

You need to purify 2.0 grams of an impure sample of Acetanilide. The sample is contaminated with aniline. After the purification

is complete you isolate 0.8 grams of acetanilide and record a melting point range of 108-110 °C. Complete the following calculations and show your work.
a. Calculate the minimum amount of distilled water you would use to complete the recrystallization.
b. How much acetanilide will still be dissolved in solution even after the sample is cooled to 0 °C?
c. Calculate the % recovery and the % error for the melting point.
d. Why is the percent recovery less than 100%? Describe multiple sources for loss of sample.

Chemistry
1 answer:
marusya05 [52]2 years ago
3 0

Answer:

Following are the answer to this question:

Explanation:

In the given question an attachment file is missing, that is attached. please find the attached file, and the following are the description of the given points:

a. At 100 degrees in 100 mL 5 g is dissolved.  

For, it required:

\to 2g = 100 \times \frac{2}{5}

         = 40 \ \ ml \ of \ water.  

b. At 0 degrees 100 mL dissolve in 0.3 g.  

So, the dissolve:

\to 40 \ ml= 0.3\times \frac{40}{100}

               = 0.12g.

After refrigeration 0.12 g will still be dissolved.  

c. After dissolving and freezing, precipitation can occur which would still be impure if the cooling is instantaneous. The added solvent was also too hard to recrystallize. The solvent was placed below its place of reservation.  

d. Recovery percentage:

\to \frac{0.8}{2}\times100

\to 40 \ \%

The melting point of acetanilide:

\to 114.3^{\circ}.

Found=109(medium)  

Melting point error percentages:

= \frac{114.3-109}{114.3}\\\\=4.63 \ \%

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Pepsi [2]
Answer : Given data ;
Δ G° = 212 KJ/molTemperature is = 25+273 = 298 KAnd gas constant R = 0.008314 KJ/mol

The reaction is NiO(s) ⇌ Ni(s) + \frac{1}{2}  O_{2 _{(g)} We can find out the pressure on oxygen by using the equation of gibb's free energy with remainder quotient which is,                                      ΔG = ΔG° +RT lnQ
when at equilibrium Q = K, here K is equilibrim constant, and ΔG becomes 0;
so we get, 0 = ΔG° + RT ln K
on rearranging we get, ln K = ΔG° / (RT)
ln K = 212 / (0.00831 X 298) = 85.6  
Here, now K =  = 1.51 X 10^{37}
When we get K = 1.51 X 10^{37}

But, K = (PO_{2})^ \frac{1}{2}So, (1.51 X 10^{37}) ^ \frac{1}{2} = 3.87 X 10^{18}
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7 0
2 years ago
A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

7 0
2 years ago
If two unidentified solids of the same texture and color have different solubilities in 100 grams of water at 20°C, you could co
AlekseyPX
I'm certain it's "D"

...because it can't be "A" or "B" because solubility IS a property but to actually determine whether these two substances are the same or different we would need at least two-three properties (like boiling point or specific heat).

and it can't be "C" because the melting point is just simply irrelevant when comparing the solubility of two substances.
3 0
2 years ago
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Answer:

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Explanation:

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Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

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BigorU [14]

Answer:

file link http/www.openfree

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