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Marat540 [252]
3 years ago
8

In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y 4 − Y4− ) and metal chelate (abbreviated MY n − 4 M

Yn−4 ) can buffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant. The equilibrium M n + + Y 4 − − ⇀ ↽ − MY n − 4 Mn++Y4−↽−−⇀MYn−4 is governed by the equation K ′ f = α Y 4 − ⋅ K f = [ MY n − 4 ] [ M n + ] [ EDTA ] Kf′=αY4−⋅Kf=[MYn−4][Mn+][EDTA] where K f Kf is the association constant of the metal and Y 4 − Y4− , α Y 4 − αY4− is the fraction of EDTA in the form Y 4 − Y4− , and [ EDTA ] [EDTA] is the total concentration of free (unbound) EDTA EDTA . K ′ f Kf′ is the conditional formation constant. How many grams of Na 2 EDTA ⋅ 2 H 2 O Na2EDTA⋅2H2O (FM 372.23 g/mol) should be added to 1.97 1.97 g of Ba ( NO 3 ) 2 Ba(NO3)2 (FM 261.35 g/mol 261.35 g/mol ) in a 500. mL volumetric flask to give a buffer with p Ba 2 + = 7.00 pBa2+=7.00 at pH 10.00? log K f log⁡Kf for Ba − EDTA Ba−EDTA is 7.88 7.88 and α Y 4 − αY4− at pH 10.00 is 0.30. mass Na 2 EDTA ⋅ 2 H 2 O.
Chemistry
1 answer:
Lyrx [107]3 years ago
5 0

Answer:

There will 3.95 grams of Na2 and H2O that should be added to form a concentric required solution.

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slava [35]

Option 3- Avogadro's, Charles's and Boyle's


6 0
3 years ago
The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of CuFeS_2 = 866.337 g = 0.8663 Kg         (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.


3 0
3 years ago
Question - Complete and balance the following chemical equations:
Mekhanik [1.2K]
Answer is because
Please give feedback
6 0
2 years ago
How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.
Tcecarenko [31]

1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

n=0.205 mole

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=200 K

Putting value in the given equation:

\frac{nRT}{P} =V

V= \frac{0.205 \;mole\;0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 200}{3 \;atm}

V= 1.1214 mL

Learn more about the ideal gas here:

brainly.com/question/27691721

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4 0
1 year ago
In the bromination of arenes, which of the following statements regarding the reaction is true?a.The hydrocarbon is used in exce
Olenka [21]

The hydrocarbon is used in excess.

<h3><u>Explanation</u>:</h3>

The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.

8 0
3 years ago
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