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chubhunter [2.5K]
2 years ago
8

Paul determines that the hydrogen ion concentration of his unknown solution is 3.60×10^-5 M. what is the pH of this solution?​

Chemistry
1 answer:
vivado [14]2 years ago
4 0

Answer:

<h2>pH = 4.44 </h2>

Explanation:

The pH of a substance can be found by using the formula

p H  =  -   log[ H^{ + }  ]

where [ H+ ] is the hydrogen ion concentration of the solution

From the question

[ H + ] = 3.60 × 10^-5 M

So the pH is

pH =  -  log(3.60 \times  {10}^{ - 5} )  \\  =4.44369749923

We have the final answer as

<h3>pH = 4.44 </h3>

Hope this helps you

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What is the difference between conjugate acid-base pair?
NeX [460]

Answer:

B. a H+ ion is the answer dear.

Explanation:

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5 0
2 years ago
A sample of CO2 gas has a volume of 2.7 L at 78.5 kPa. At what pressure would this sample of gas have a volume of 4.0L? Temperat
lisabon 2012 [21]

Answer:

52.99 kPa

Explanation:

Initial volume V1 = 2.7 L

Initial Pressure P1 = 78.5 kPa

Final Volume V2 = 4.0L

Final Pressure P2 = ?

Temperature is constant

The relationship between these quantities is given by the mathematical expression of Boyles law. This is given as;

V1P1 = V2P2

P2 = V1P1 / V2

P2 = 2.7 * 78.5 /  4.0

P2 = 52.99 kPa

7 0
3 years ago
How many water molecules self ionize in one liter of water
Drupady [299]

Answer:

yes

Explanation:

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5 0
3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
3 years ago
The wavelength of a particular color of violet light is 433 nm. The energy of this wavelength of light is kJ/photon. (109 nm = 1
mash [69]

Answer:

4.59 × 10⁻³⁶ kJ/photon

Explanation:

Step 1: Given and required data

  • Wavelength of the violet light (λ): 433 nm
  • Planck's constant (h): 6.63 × 10⁻³⁴ J.s
  • Speed of light (c): 3.00 × 10⁸ m/s

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

433 nm × 1 m/10⁹ nm = 4.33 × 10⁷ m

Step 3: Calculate the energy (E) of the photon

We will use the Planck-Einstein's relation.

E = h × c/λ

E = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/4.33 × 10⁷ m

E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ

5 0
2 years ago
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