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trapecia [35]
3 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]3 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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Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry.
KiRa [710]

Answer:

A) 54.04%

B) 13-karat

Explanation:

A) From the problem we have

<em>1)</em> Mg + Ms = 9.40 g

<em>2)</em> Vg + Vs = 0.675 cm³

Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.

We can rewrite the first equation using the density values:

<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40

So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:

We <u>express Vg in terms of Vs</u>:

  • Vg + Vs = 0.675 cm³
  • Vg = 0.675 - Vs

We <u>replace the value of Vg in equation 3</u>:

  • Vg * 19.3 + Vs * 10.5 = 9.40
  • (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
  • 13.0275 - 19.3Vs + 10.5Vs = 9.40
  • -8.8 Vs + 13.0275 = 9.40
  • <u>Vs = 0.412 cm³</u>

Now we <u>calculate Vg</u>:

  • Vg + Vs = 0.675 cm³
  • Vg + 0.412 cm³ = 0.675 cm³
  • Vg = 0.263 cm³

We <u>calculate Mg from Vg</u>:

  • 0.263 cm³ * 19.3 g/cm³ = 5.08 g

We calculate the mass percentage of gold:

  • 5.08 / 9.40 * 100% = 54.04%

B)

We multiply 24 by the percentage fraction:

  • 24 * 54.04/100 = 12.97-karat ≅ 13-karat
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Answer:

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Explanation:

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2 years ago
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A gas has a density of 1.57 g/L at 40.0 °C and 2.00 atm of pressure. What is the identity of the gas?
Naddika [18.5K]

Answer:

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Explanation:

Step 1: Given and required data

  • Density of the gas (ρ): 1.57 g/L
  • Temperature (T): 40.0°C
  • Pressure (P): 2.00 atm
  • Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

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