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Snezhnost [94]
3 years ago
11

what is the number of each type of atom on the right side of the equation 2Na3PO4(aq)+2CoCl2(aq)→2Co3(PO4)2(s)+6NaCl(aq)

Chemistry
2 answers:
Luden [163]3 years ago
7 0

Answer : The number of atoms Co,P,O,Na\text{ and }Cl are, 6, 4, 16, 6, 6

Explanation :

The given balanced chemical reaction is:

2Na_3PO_4(aq)+2CoCl_2(aq)\rightarrow 2Co_3(PO_4)_2(s)+6NaCl(aq)

This reaction is balanced reaction because in this reaction all the atoms of individual elements are completely balanced.

From the balanced reaction we conclude that Co_3(PO_4)_2 and NaCl are present on the right side of the equation.

Number of atoms of cobalt (Co) present on right-side of the equation = 6

Number of atoms of phosphorous (P) present on right side of the equation = 4

Number of atoms of oxygen (O) present on right side of the equation = 16

Number of atoms of sodium (Na) present on right side of the equation = 6

Number of atoms of chlorine (Cl) present on right side of the equation = 6

Hence, the number of atoms Co,P,O,Na\text{ and }Cl are, 6, 4, 16, 6, 6

Andre45 [30]3 years ago
3 0

6 atoms of Cobalt.

4 atoms to Phosphate (PO4).

6 atoms to sodium.

6 atoms to Chlorine.

The coefficient is one factor that tells you how many atoms go to each atom/element, however, the subscript also influences this.

If the compound has parenthesis, then the subscript within the parenthesis remains untouched and does not affect the atoms.

Basically, to calculate the atoms is multiply the coefficient (number in front of the atom) by the subscript attached to the atom. If it's a compound such as Co3PO4, then make sure the coefficient affects the second atom as well.

Hope this helps!

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Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

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  • O = 2

Number of atoms present on product side are as follows.

  • C = 1
  • H = 2
  • O = 3

To balance this equation, multiply O_{2} by 2 on reactant side. Also, multiply H_{2}O by 2 on product side. Hence, the equation can be rewritten as follows.

CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O

Now, the number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 4

Number of atoms present on product side are as follows.

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  • O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

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Explanation:

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2) Aspartate contains an extra carboxyl group with a dissociable protron. Once the Protron is dissociated, it carries an extra negative charge in its side chain (R)

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