Ask question

Ask question

All categories

erastovalidia [21]

2 years ago

12

In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th

ree times with 5.0-ml portions of methylene chloride. calculate the total amount of caffeine that can be extracted into the three portions of methylene chloride (see technique 12, section 12.2). caffeine has a distribution coefficient of 4.6 between methylene chloride and water.

1 answer:

zmey [24]2 years ago

6 0

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

$w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms$

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

=0.045 gms

You might be interested in

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C

UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M 0.24 M 0.48 M 0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0

2 years ago

How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a

Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

Hope this helped :)

5 0

2 years ago

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t

TEA [102]

Explanation:

The reaction equation will be as follows.

$Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

= $50 \times 0.6$

= 30 mmol

Similarly, calculate the number of moles of HBr as follows.

No. of moles of HBr = $M \times V$

= $50 \times 0.6$

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) = $m \times s \times \Delta T$

as, m = mass of solution

and, Density = $\frac{mass}{volume}$

or, mass = Density × Volume

= 1.08 g/ml \times (50 + 50) ml

= 108 g

where, s = specific heat of solution = 4.18 j/g.k

and, change in temperature $\Delta T$ = $(26 - 23)^{o}C$

= $3
^{o}C$

Hence, the heat released will be as follows.

q = $m \times s \times \Delta T$

q = $108 \times 4.18 \times 3^{o}C$

= 1354.32 joule

or, = 1.354 kJ (as 1 kJ = 1000 J)

Also, $\Delta H_{rxn}$ = $\frac{-q}{n}$

= $\frac{-1.354}{15 \times 10^{-3}}$

= -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

6 0

2 years ago

Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O

oee [108]

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

<u>To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺</u>:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the <u>number of Cr atoms</u> on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, <u>6 electrons are gained</u> by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to <u>balance the total charge</u> on the reactant and product side,<u> 14 H⁺ is added on the reactant side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to <u>balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

<u>Therefore, the correct balanced reduction half-reaction is:</u>

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

3 0

2 years ago

Neither dry soil nor pure water conducts electricity. But wet soil will conduct electricity. Explain why this happens?

kipiarov [429]

Pure water does not have enough ions to conduct electricity. A mixture of metals such as iron, zinc and copper in the wet soil can trigger electrolysis that requires excess energy in the form of over potential to conduct electricity. The excess energy is needed due to limited self-ionization of water. The wet soil then can conduct current when positive and negative ions are present. The water ions begin to flow from anode (positive electrode) to cathode (negative electrode) to be oxidize and produce electricity.

<span> </span>

4 0

2 years ago