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belka [17]
3 years ago
13

Which of the following is TRUE? a) A neutral solution contains [H2O] = [H3O⁺] b) A neutral solution does not contain any H3O+ or

OH- c) An acidic solution has [H3O⁺] > [OH⁻] d) A basic solution does not contain H3O+ e) None of the above are true.
Chemistry
1 answer:
Tpy6a [65]3 years ago
3 0
<h2>An acidic solution has [H_3O^+] > [OH^-] </h2>

Explanation:

pH is the negative logarithm of hydronium ion concentration present in a solution.

If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9. The solution has more [H_3O^+] than [OH^-]

If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14. The solution has more [OH^-]  than [H_3O^+]

The solution having pH equal to 7 is termed as neutral solution.The solution has equal concentration of [H_3O^+] and [OH^-]

Learn more about acidic, basic and neutral solution

brainly.com/question/13196724

brainly.com/question/13155459

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Answer: % Errror = 11.05%

Explanation:

Percentage error is the difference between the measured value and the theoretical value as expressed in percent.

That is;

% Error =

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Theoretical value =55.2kJ

Measured value = 49.1kJ

% Error = (55.2-49.1)/(55.2) ×100

%Error = 6.1/55.2 ×100

% Error = 0.1105 ×100

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"benzyl ethyl ether reacts with concentrated aqueous hi to form two initial organic products (a and b). further reaction of prod
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Answer:

See below.  

Step-by-step explanation:

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R-OR' + HI ⟶ R-I + H-OR'

<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.

  1. PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻    Protonation of the ether
  2. PhCH₂-O⁺(H)R  ⟶ PhCH₂⁺ + HOR          Sₙ1 ionization of oxonium ion
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If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:

ROH +HI ⟶ R-I + H-OH

Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).

The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).

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