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2 years ago

Consider the following combustion reaction:

Chemistry

2 answers:

borishaifa [10]2 years ago

8 0

Answer: 0.34827 mole CO2.

Explanation: Solved in the attached picture.

ZanzabumX [31]2 years ago

6 0

Answer:

vxcvggccbbcCbnmhxx vgfscvjjg

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20. In which medium do sound waves move the fastest?

Nataliya [291]

Answer:

D. Solids

Explanation:

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2 years ago

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What two properties of a gas depend on its container?

Bogdan [553]

One property is it's volume. I am not sure if the second

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3 years ago

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How is sodium extracted from is ore?

Aloiza [94]

Answer:

Sodium is extracted from it's ore by electrolysis of fused sodium chloride.

Explanation:

he process is usually carried out ia a special electrochemical cell called the downs cell. While molten sodium metal is collected at the cathode and also sent to tanks for cooling and storage.

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3 years ago

What is the classification of a solution of naoh with a ph of 8.3?

Maksim231197 [3]

Answer:

<u>Alkaline or basic solution </u>(alkaline and basic means the same)

Explanation:

According to the <em>pH</em>, solutions may be classified as neutral, acidic, or alkaline (basic).

This table shows such classification:

pH classification

7 neutral

> 7 alkaline or basic

< 7 acidic

Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).

Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.

pH = - log [H₃O⁺]

You can calculate the concentration of hydronium ions using antilogarithm properties:

$pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501$

NaOH solutions are alkaline solutions, bases, according to Arrhenius model, because they contain OH⁻ ions and release them when ionize in water.

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3 years ago

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

soldi70 [24.7K]

Answer: from the Zn anode to the Cu cathode

Justification:

1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)

2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).

3) Then, you can already tell that electrons go from Zn to Cu.

4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.

So you get that the electrons flow from the anode (Zn) to the cathode (Cu).

Always oxidation occurs at the anode, and reduction occurs at the cathode.

3 0

3 years ago