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hram777 [196]
3 years ago
6

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Chemistry
1 answer:
____ [38]3 years ago
7 0

Answer:

ertyuioekgdctifrofy?

Explanation:

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What is a polyprotic buffer?
coldgirl [10]
Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different
3 0
2 years ago
An organic compound has a distribution coefficient of 1.5
jasenka [17]

Answer:

Three times with 5 ml will yield more

Explanation:

Let x represent the amount yield

Kd = (x/15) / ((50-x) / 15) where Kd = 1.5

1.5 = (x/15) / ((50-x) / 15)

x / (50 - x) = 1.5

x = 75 - 1.5x

x + 1.5x  = 75

2.5x = 75

x = 75 / 2.5 = 30 mg

when extraction  three times

1st extraction

(x1/5) / ((50 - x1) / 15) = 1.5

3x1 / 50 - x1 = 1.5

3x1 = 75 - 1.5x1

3x1 + 1.5x1 = 75

4.5x1 = 75

x1 = 75 / 4.5 = 16.67 gm

second extraction

(x2/ 5) / (33.33 - x2 ) / 15) = 1.5

3x2 / ( 33.33 - x2) = 1.5

3x2 = 1.5(33.33 - x2)

3x2 = 49.995 - 1.5x2

3x2 + 1.5x2 = 49.995

4.5x2 = 49.995

x2 = 49.995 / 4.5

x2 = 11.11 mg

Third extraction

(x3/5) / ((22.22 - x3) / 15) = 1.5

3x3 = 1.5 ( 22.22 - x3)

3x3 + 1.5x3 = 33.33

4.5x3 = 33.33

x3 = 33.33 / 4.5 = 7.41 mg

total extraction = x1 + x2 + x3 =16.67 + 11.11 + 7.41 = 35.19 mg

the three times extraction using 5ml yields 5.19 mg more

3 0
2 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
Can anyone do these? I have 20 points for who ever can get them. I've completed everything else and these ones are what I'm stuc
8_murik_8 [283]
It would be something nvm forgot it
8 0
3 years ago
What are the examples of all the non metals in the first twenty element​
agasfer [191]

Answer:

semimetals or metalloids.

Explanation:

5 0
3 years ago
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