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sdas [7]
3 years ago
11

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form.

Chemistry
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:- \frac{[In^-]}{[HIn]}=3.16*10^-^8

Solution:- If both acid and basic form are present at the same time then it's like a buffer solution since the buffer solution also has a weak acid and its conjugate base. The Handerson equation is used for solving buffer problems. Let's use the same equation here also.

The equation is:

pH=pKa+log(\frac{base}{acid})

pKa is given as 9.7 and the pH is 2.2. let's plug in the values in the equation.

2.2=9.7+log(\frac{[In^-]}{[HIn]})

subtract 9.7 from sides:

2.2-9.7=log(\frac{[In^-]}{[HIn]})

-7.5=log(\frac{[In^-]}{[HIn]})

taking antilog:

10^-^7^.^5=\frac{[In^-]}{[HIn]}

3.16*10^-^8=\frac{[In^-]}{[HIn]}

So, the answer is 3.16*10^-^8 .

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Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) -7.3
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Answer:

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Pump work (W) is calculated as

W = (h_f - h_i) \times \dot{m}

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h_i is the enthalpy of water at its initial state

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3 years ago
Use the rhombus to find the measure of the following angles m
Studentka2010 [4]

DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°. This can be obtained by understanding the properties of a rhombus.

<h3>What are the required values of sides and angles?</h3>

Here in the question it is given that,

  • ABCD is a rhombus
  • DB = 10, BC = 13, ∠WAD = 25°

We have to find the values of DA, BW, WC, ∠BAC, ∠ACD, ∠DAB, ∠ADC, ∠DBC, ∠BWC.

1) Since all the sides of the rhombus are equal, DA = 13

2) Since diagonals are perpendicular bisectors of each other, BW = 5

3) Since formula of diagonal is,

d₁ = √4a² - d₂,

here a = 13, d₂ = 10

d₁ = √4×13² - 10

d₁ = √676 - 10 = √576 = 24

Since diagonals are perpendicular bisectors of each other

WC = 24/2  ⇒ WC = 12

4) Since diagonals are angle bisectors at the corners, ∠BAC = 25°

5) Since diagonals are angle bisectors at the corners,

∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

Since opposite angles of the rhombus are equal,

∠BAD = ∠BCD = 50° ⇒ ∠BCW = ∠DCW = 25°

⇒ ∠ACD = 25°

6) Since diagonals are angle bisectors at the corners,∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

⇒ ∠DAB = 50°

7) Since adjacent angles are supplementary angles,

∠BAD + ∠ADC = 180° ⇒ 50° + ∠ADC = 180° ⇒ ∠ADC = 180° - 50° = 130°

⇒ ∠ADC = 130°

8) Since diagonals are angle bisectors at the corners,

∠DBC = 130°/2

⇒ ∠DBC = 65°

9) Since all the sides make an angle of 90° at the center,

⇒ ∠BWC = 90°

Hence DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°.

Learn more about rhombus here:

brainly.com/question/27870968

#SPJ9

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