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kifflom [539]
3 years ago
13

A solution has a pH of 11.75. Calculate the [H:0*] of the solution

Chemistry
1 answer:
Savatey [412]3 years ago
7 0

The [H₃0⁺] of the solution : 1.778  x 10⁻¹² M

<h3>Further explanation  </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.  

pH = - log [H⁺]  

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.  

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.  

pH of solution = 11.75

the [H₃0⁺] of the solution :

\tt pH=-log[H_3O^+]\\\\11.75=-log[H_3O^+]\\\\(H_3O^+]=10^{-11.75}=1.778\times 10^{-12}

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2. In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] =
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3 0
3 years ago
What is the percentage of oxygen in Li(NO2)3
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Answer:

66.2 % of O

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Our compound is the lithium nitrite.

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This salt is ionic and can be dissociated: LiNO₂ →  Li⁺ + NO₂⁻

We determine the molar mass:

molar mass of Li + 3 . molar mass of N + 6 . molar mass of O

6.94 g/mol + 3. 14 g/mol + 6 . 16 g/mol = 144.94 g/mol

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3 0
3 years ago
What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi
Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an <em>exothermic reaction</em>. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, <em>Quantity of heat, q = mass * specific heat capacity * temperature rise.</em>

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

6 0
3 years ago
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