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Dafna1 [17]
3 years ago
11

Balance N2 + H2 —NH3 what type of reaction

Chemistry
1 answer:
AnnyKZ [126]3 years ago
5 0
3! You have to ensure balance of all the different elements.
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The _________________ are either elements or compounds which are present prior to a chemical reaction.
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Reactants would be the missing word in the sentence because they make up the products in a chemical reaction
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The product side of a chemical reaction is shown.
Inga [223]

The number that represents the coefficient on the product side of the chemical reaction, -- > 7Ti_2(SO_4)_3 is 7.

<h3>Coefficients of chemical equations</h3>

In equations representing chemical reactions, the coefficient of each reactant or product of a reaction is the number that comes on the left-hand side just before the chemical formula.

The coefficient of each species in a chemical reaction is obtainable when the equation of the reaction is balanced.

For example, in the following equation: 2A + B = 3C + D

The coefficients of A, B, C, and D are 2, 1, 3, and 1 respectively.

Applying this to the product side of a chemical reaction;  -- > 7Ti_2(SO_4)_3

It means that the coefficient of the product is 7.

More on coefficients of chemical equations can be found here: brainly.com/question/28294176

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7 0
1 year ago
Which of the following correctly describes what happens to the kinetic energy of water when it changes from steam to a liquid?
asambeis [7]

Answer:

The kinetic energy of the molecules increases

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5 0
2 years ago
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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