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Burka [1]
4 years ago
13

The kinetic energy of all the particles in a given sample of matter is the different. Question 11 options: True False

Chemistry
1 answer:
GalinKa [24]4 years ago
6 0

Answer is true I'm voting

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Evaporation because the salt would dissolve in the water then boil the water and it will evaporate leaving the salt left in the bowl
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3 years ago
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Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu
borishaifa [10]

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

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4 years ago
How can toxicant absorption be reduced after exposure to the skin? How can absorption be reduced for orally consumed chemicals?
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Answer:

The toxicant absorbtion can be reduced after exposure to the skin the surrounding clothing shoes or gloves should be removed or torn off than the part of the body which was exposured to the toxicant should be immediatly washed using clean running water for a while, with cold shower being the most recommended splashing method. In case of absorption for orally consumed chemicals should check on any remaining toxicant residue and be removed from the mouth. Vomiting should be induced to patients that are still conscious by providing them with liquids that can provoke vomiting. This will help in removing the toxicant in the intestinal and reduce their effect. Gastric lavage should then be done to induce diarrhea.  

Explanation:

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Write a balanced equation for the oxidation-reduction reaction that occurs when hydrogen peroxide reacts with ferrous ion
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H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O

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Fe²⁺ → Fe³⁺ + e⁻                 k=2

H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺


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