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olya-2409 [2.1K]
3 years ago
8

To run a thin layer chromatography experiment with a chemical substance, begin by marking a horizontal line near the bottom of a

TLC plate with_______. Place a_______spot of the substance onto the line. For the mobile phase, add a small amount of _______ at the bottom of a TLC chamber. Place the plate in, then_______the chamber. Once the mobile phase approaches the top of the plate, remove the plate and mark the _______ line. Note the position of the spots and calculate Rf values if needed.
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
8 0

Answer:

TLC is thin-layer chromatography, a chromatography technique which is used for separating the non-volatile mixtures.

Explanation:

To run a thin layer thin layer chromatography experiment with a chemical substance, begin by marking a horizontal line near the bottom of TLC plate with PENCIL. Place a SMALL spot of the substance onto the line. For the mobile phase add a small amount of SOLVENT at the bottom of TLC chamber. Place the plate in, then COVER the chamber. Once the mobile phase approaches the top of the plate, remove the plate and mark the SOLVENT line. Note the positions of the spot and calculate the Rf if needed.

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A hiker has packed a bag of chips as a snack. At the start of the hike the pressure was 1.5 atm and the temperature was 35*C. At
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According to <span>Gay-Lussac's Law the temperature and Pressure are directly proportional to each other if the amount and volume of given gas are kept constant.
Mathematically for initial and final states it is expressed as,

                                          P</span>₁ / T₁  =  P₂ / T₂     ----- (1)
Data Given;
                  P₁  =  1.5 atm

                  T₁  =  35 °C + 273  =  308 K

                  P₂  =  ?

                  T₂  =  0 °C + 273  =  273 K

Solving Eq. 1 for P₂,

                                   P₂  =  P₁ T₂ / T₁

Putting values,
                                   P₂  =  (1.5 atm × 273 K) ÷ 308 K

                                   P₂  =  1.32 atm
Result:
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5 0
3 years ago
It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L

1 m^3=1000 l

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%

Total mass of hydrogen hexafluorosilicate = m'

79.16\%=\frac{25,434 g}{m'}\times 100

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

5 0
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zavuch27 [327]

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Explanation:

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i = van’t Hoff factor

Pi = iMRT

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7 0
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Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize t
almond37 [142]

Answer:

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

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Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole

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percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%

= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

3 0
3 years ago
What phenomenon occurs after mass extinction?
vampirchik [111]

Answer:

Diversification and adaptive radiation.

Explanation:

3 0
3 years ago
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