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Elan Coil [88]
3 years ago
13

Scientists notice that many fish are dying in a local lake. Which onsite water quality data would they most likely gather to det

ermine the cause of the fish kill?
A. arsenic and cyanide levels

B.water level and surface area

C.biodiversity and water depth

D. dissolved oxygen and temperature
Chemistry
1 answer:
never [62]3 years ago
8 0

Answer:

D. (Dissolved oxygen and temperature)

Explanation:

Have a nice day and stay safe <3

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It has 2 valence electrons.

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Is dynamite an atom or a molecule?
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TNT is a molecule

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What is true about protons within an atom?
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Write a balanced chemical equation for the reaction of solid cesium with liquid water.
Alborosie

The balanced equation for reaction of solid cesium with liquid water

= 2Cs + 2H2O → 2CsOH + H2

cesium react with liquid water to produce cesium hydroxide and hydrogen gas

that is 2 moles of Cs react 2 moles of H2O to form 2 moles CsOH and 1 of hydrogen gas

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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

6 0
3 years ago
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