HELP I DON'T HAVE LONG LEFT AND I'M STRUGGLING SO BAD PLEASE I BEG U TO HELP

Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:

Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O

After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:

(19mg - X)

<em>Where X is the mass that now is in the aqueous phase</em>

Replacing in Kp formula:

9.5 = (19mg - X) / 5mL / (X /10mL)

0.95X = 19mg - X / 5mL

4.75X = 19 - X

5.75X = 19

X = 19 / 5.75

X = 3.30mg

That means 9-fluorenone that remain in the ether layer is:

19mg - 3.30mg =

<h3>15.70mg would remain</h3>

8 0

3 years ago

How many grams of methionine (MW = 149.21) are needed to make 20 mL of a 150 mM solution?

Leni [432]

Answer:

0.45 g

Explanation:

Step 1: Given data

Molar mass of methionine (M): 149.21 g/mol

Volume of the solution (V): 20 mL

Concentration of the solution (C): 150 mM

Step 2: Calculate the moles of methionine (n)

We will use the following expression.

n = C × V

n = 150 × 10⁻³ mol/L × 20 × 10⁻³ L

n = 3.0 × 10⁻³ mol

Step 3: Calculate the mass of methionine (m)

We will use the following expression.

m = n × M

m = 3.0 × 10⁻³ mol × 149.21 g/mol

m = 0.45 g

7 0

3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago