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vazorg [7]
3 years ago
15

How many moles of coF3 are produced when 1.25 moles of 3co reacts

Chemistry
2 answers:
oee [108]3 years ago
7 0

Answer:

1.25 moles

Explanation:

First, we need to balance the equation. Essentially, this means making sure we have the same number of each atom on each side.

On the left side, we currently have:

- 1 Co atom

- 2 F aomts

On the right side, we have:

- 1 Co atom

- 3 F atoms

To balance it, add a 2 to Co on the left, 3 to F2 on the left, and 2 to CoF3 on the right:

2Co+3F_2 → 2CoF_3

Now, we have 1.25 moles of Co, and since the ratio between Co and CoF3 is 1:1, we also have 1.25 moles of CoF3.

Thus, the answer is 1.25 moles.

Sloan [31]3 years ago
3 0

Answer:

1.25 moles

Explanation:

Balancing the equation:

2Co + 3F2 --> 2CoF3

Co : CoF3

2 : 2

1.25 : X

X/1.25 = 2/2

X/1.25 = 1

X = 1.25 moles

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If you add oil to water and shake the two liquids together what will you form
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The equation for the reaction between copper and nitric acid is vCu + wHNO3--->xCu(NO3)2 + yNO + zH2O Balance the equation an
lana [24]

Answer:

The coefficients of the balanced equation are:

  • v= 1
  • w= 4
  • x= 1
  • y= 2
  • z=2

Explanation:

The Law of Conservation of Matter, also called the Law of Conservation of Mass or the Lomonósov-Lavoisier Law, postulates that "mass is neither created nor destroyed, it is only transformed". This means that the reagents interact with each other and form new products with different physical and chemical properties than the reagents, but the amount of matter or mass before and after a transformation (chemical reaction) is always the same. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

Then, you must balance the following chemical equation:

v Cu + w HNO₃ → x Cu(NO₃)₂ + y NO + z H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

  • Left side:   1 copper, 1 hydrogen, 1 nitrogen and 3 oxygen.
  • Right side:  1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Generally, hydrogen and oxygen balance in the end. So you balance nitrogen first, because copper is already balanced (there is the same amount on both sides of the reaction). If w = 2, then:

  • Left side:   1 copper, 2 hydrogen, 2 nitrogen and 6 oxygen.
  • Right side:  1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

But you see then that the oxygen is unbalanced and you have less quantity in the reagents. Then w must be greater. Being w = 4:

  • Left side:   1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
  • Right side:  1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

Going back to the idea of ​​balancing nitrogen, being y = 2:

  • Left side:   1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
  • Right side:  1 copper, 2 hydrogen, 4 nitrogen and 9 oxygen.

Balancing the hydrogen, being z = 2:

  • Left side:   1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
  • Right side:  1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.

Since you have the same amount of each element on each side of the reaction, the reaction is balanced. Then, the balanced equation is:

Cu + 4 HNO₃ → Cu(NO₃)₂ +  2 NO + 2 H₂O

Finally:

  • <u><em>v= 1</em></u>
  • <u><em>w= 4</em></u>
  • <u><em>x= 1</em></u>
  • <u><em>y= 2</em></u>
  • <u><em>z=2</em></u>
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Answer:

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Explanation:

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Explanation:

The natural gas consists of chemicals, which when burn produces thermal energy. This thermal energy is utilized for heating or boiling water. The boiling water produces steam the steam tries to come out of the kettle thus makes sound over the kettle lid, the pressure created by steam while leaving kettle produces sound energy.

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