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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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I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please
lara31 [8.8K]

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

8 0
2 years ago
Identify 5 basic examples of Solutions?
Elden [556K]
Coffe
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A solution is any liquid
7 0
3 years ago
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Which of the following is the best definition of an independent variable
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Answer:

that one

Explanation:

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8 0
3 years ago
27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a
jok3333 [9.3K]

Answer : The volume of the cube is, 21.88cm^3

Solution : Given,

Density of nickel = 8.91g/cm^3

Number of nickel atoms = 2\times 10^{24}

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, 6.022\times 10^{23} atoms form 1 mole of nickel

So, 2\times 10^{24} atoms form \frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321 moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}

\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

Density=\frac{Mass}{Volume}

8.91g/cm^3=\frac{194.94g}{Volume}

Volume=21.88cm^3

Therefore, the volume of the cube is, 21.88cm^3

4 0
3 years ago
Which element is in the same family as Chlorine (Cl) and Fluorine (F)?
ValentinkaMS [17]

Answer/Explanation:

Chlorine and Fluorine are in the Halogen family. The elements in the Halogen family are:

Fluorine (F)

Chlorine (Cl)

Bromine (Br)

Iodine (I)

Astatine (At)

Tennessine (Ts)

Hydrogen (H) is a nonmetal

Oxygen (O) is a nonmetal

Lithium (Li) is an alkaline metal.

8 0
3 years ago
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