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vladimir2022 [97]
3 years ago
12

hich of the following statements is not true for an exothermic reaction? Question 2 options: The products have a higher heat con

tent than the reactants. The temperature of the reaction system increases. The temperature of the surroundings increases. Heat passes from the reaction system to the surroundings. The enthalpy change for the reaction is n
Chemistry
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.

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Write the formula for the polyatomic ion in KOH. Express as an ion..
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KOH is a compound containing two ions, K+ and OH-. 
<span>The polyatomic ion present is OH- which is called hydroxide. </span>
<span>The compound is named potassium hydroxide.</span>
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Forces that change an object motion by touching the object are _______ forces
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25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?
weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

CH_3COOH + NaOH ->  CH_3COONa + H_2O

First step. Need to know how much moles of the substances are present

0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> Na^{+} + OH^{-}

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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Answer: bubbles form in boiling water

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