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ololo11 [35]
2 years ago
14

What is the hydronium ion concetration of a ph=9. 2 solution

Chemistry
1 answer:
serg [7]2 years ago
3 0

The hydronium ion concentration of a pH = 9.2 of solution is 6.3 × 10⁻¹⁰M.

<h3>How do we calculate pH?</h3>

pH of the solution is define as the negative logarithm of the concentration of the hydronium ion as:

pH = -log[H⁺]

Or [H⁺] = 10^{-pH}

Given that pH value = 9.2

On putting values, we get

[H⁺] = 10^{-9.2} = 6.3 × 10⁻¹⁰M

Hence required concentration of hydronium ion is 6.3 × 10⁻¹⁰M.

To know more about pH, visit the below link:

brainly.com/question/172153

#SPJ2

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Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (
Nesterboy [21]

Balanced chemical equation for the reaction is:

2SO_{2} (g) + O_{2} (g)+ 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

Moles of H_{2} S_{} O_{4} formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of SO_{2} = 11.5 moles

moles of H_{2} S_{} O_{4} = ?

Moles of O_{2} needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2SO_{2}(g) + O_{2} (g) + 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of SO_{2}, x moles of H_{2} S_{} O_{4} is formed

\frac{1}{2}  =\frac{x}{11.5}

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1  mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

7 0
4 years ago
The higher the pH, the less acidic the solution
marshall27 [118]

Answer:

<em>yh thats true lol, ty for that very interesting fact</em>

6 0
3 years ago
Read 2 more answers
A scientific law is ___.
Daniel [21]

Answer: a rule that describes a pattern in nature.

Explanation:

3 0
3 years ago
if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was redu
Nata [24]

Answer:

1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

1 × 1.5 = P2 × 0.8

1.5 = P2 × 0.8

Divide both side by 0.8

P2 = 1.5/0.8

P2 = 1.875 atm

Finally, we shall convert 1.875 atm to mmHg.

This can be obtained as follow:

1 atm = 760 mmHg

Therefore,

1.875 atm = 1.875 × 760 = 1425 mmHg.

Therefore, the new pressure of the gas is 1425 mmHg.

3 0
3 years ago
Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation
nikdorinn [45]

<u>Answer:</u> The isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

<u>Explanation:</u>

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is _{56}^{139}\textrm{Ba}

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is _{38}^{89}\textrm{Sr}

Hence, the isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

7 0
3 years ago
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