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8_murik_8 [283]
3 years ago
7

How does oxidation differ from reduction

Chemistry
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

See the explanation below, please.

Explanation:

In redox reactions (oxide reduction) the transfer of electrons between compounds occurs.

 In oxidation, electrons are lost (positive charge increases). Example:

Mg ---> Mg2 + + 2 e- (passes from Mg ° to Mg 2+)

On the other hand in the reduction, electrons are gained (negative charge increases or the positive one increases). Example:

Cl2 + 2e- ---> 2 Cl- + 2 e- (passes from Cl ° to Cl-)

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Stolb23 [73]

Answer:

.14L or 140mL

Explanation:

This is a classic plug-n-chug problem. Your textbook probably goes over this formula as M_{1}V_{1} = M_{2}V_{2}. M stands for molarity of the given substance, and V stands for the volume that the substance occupies.

Simply plug in the values that you're given, like so:

M_{1} = 1.5M\\V_{1} = ???\\M_2 = 7M\\V_2 = .03L

1.5M * x = 7M * .03L

After completing the algebra portion and solving for the unknown, you will be left with x = .14L, which is the volume required to neutralize 30mL of 7M NaOH.

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3 years ago
What is the empirical formula of a compound that breaks down into 4.12g of N and 0.88g of H? N 5 H NH 3 NH 4 N 4 H
Marysya12 [62]

<u>The </u><u>emperical formula </u><u>can be </u><u>calculated </u><u>as follows:</u>


<u>mass :</u>   4.12g of N            and        0.88g of H

<u>moles:</u>   4.12g/14.007       and        0.88g/1.008

               

<u>moles:</u>   0.294                   and        0.873


<u>moles/smallest number</u>:   (gives the ratio of atoms)


             0.294/0.294                and           0.873/0.294

                   1                               and             2.969

                    1                              and              3



The simplest ratio of N : H is 1 : 3

<u>Hence the </u><u>emperical formula </u><u>is </u><u>NH3</u>


<u>Explanation:</u>

 1.   The emperical formula can be found by getting  the <u>simplest ratio of atoms in a molecule.</u>

2. To get the simplest ratio of atoms we find out the <u>number of moles of each atoms in it.</u>

3. From the number of moles of atoms we can find the simplest ratio of each atom by dividing the <u>number of moles by the smallest number.</u>


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