When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nacl solution. explain why?

Answers

Answer 1

Answer:

Solution:

After the reaction of mixture is worked-up Washing three times the organic with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

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You have 0.500 L of an 0.250 M acetate buffer solution (i.e. [HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M) at pH 3.50. How many mL of 1.000 M NaOH must you add in order to change the pH to 5.07? Acetic acid has a pKa of 4.74.

Answers

Answer:

80mL of 1.00M NaOH

Explanation:

Using H-H equation, we can determine oH of a buffer as acetate buffer. First, we need to determine amount of acetate ion and acetic acid at pH 3.50 and 5.07. Then, with the reaction of NaOH with acetic acid we can find the amount of 1.00M NaOH that must be added:

At pH 3.50:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

3.50 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

0.057544 = [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M

[HC₂H₃O₂] + 0.057544[HC₂H₃O₂] = 0.250 M

1.057544 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.2364M * 0.500L = 0.1182 moles of acetic acid at first pH

At pH 5.07:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

5.07 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

2.13796= [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + 2.13796[HC₂H₃O₂] = 0.250 M

3.13796 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.07967M * 0.500L = 0.0398 moles of acetic acid at first pH

Now, NaOH reacts with HC₂H₃O₂ as follows:

NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O

As moles of acetic acid decreases from 0,1198 moles - 0,0398 moles = 0,08 moles of acetic acid are consumed = 0,08 moles of NaOH

0,08 mol NaOH * (1L / 1mol) = 0,08L of 1.00M NaOH =

80mL of 1.00M NaOH

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Answers

Answer:

Abe chutiye, Question sahi se likh laude phle

Calculate the wavelength of A 75 kg athlete running a 7.0-minute mile

Answers

Answer:

$\lambda =2.31x10^(-36)m$

Explanation:

Hello,

In this case, since the Broglie's wavelength for bodies is defined via:

$\lambda =(h)/(mv)$

Whereas h accounts for the Planck's constant, m the mass and v the velocity, which is:

$v=(1mile)/(7.0min)*(1609.34m)/(1mile)*(1min)/(60s)=3.83(m)/(s)$

Thus, the wavelength turns out:

$\lambda =(6.63x10^(-34)kg(m^2)/(s) )/(75kg*3.83(m)/(s) ) \n\n\lambda =2.31x10^(-36)m$

Best regards.

Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

$2I^-(aq)---- > I_2(s)+2e^-$

Half reaction for the cell:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

$2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq)---- > I_2(s)+2e^-$ , Reduction potential is 0.53V

Reduction half reaction: $ClO_2(g)+e^----- > ClO_2^-$ ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

$2I^-(aq)---- > I_2(s)+2e^-$

Find more information about Reduction potential here:

brainly.com/question/7484965

Answer: The half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

$2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V$

Reduction half reaction: $ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V$ ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

What units of measurement would you use to measure height and mass

Answers

Answer:

the awnser is kilograms(Kg)

PLZZZZZZZZZZZZ HELPPPPPPPPP BRAINLIEST FOR WHO GETS IT RIGHTTTTTTWhat is the mass of reactants in the following equation?N2 + 3H2 ----> 2NH3
Question 3 options:

34.05 amu

31.03 amu

30.02 amu

15.01 amu

Answers

Answer: 34.05

Explanation:

2N and 6H = abt 34

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