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Otrada [13]
3 years ago
12

Based on the replacement reaction, what would the products of the reaction be?

Physics
2 answers:
lidiya [134]3 years ago
5 0

Answer:

OH Would be the answer.

Explanation:

got it correct on edg

cricket20 [7]3 years ago
4 0

Answer:

\rm Be(OH)_2 and \rm (NH_4)_2 SO_4. The missing ion would be \rm OH^{-}.

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

  • \rm BeSO_4, and
  • \rm NH_4 OH.

In particular,

  • \rm BeSO_4 is made up of \rm Be^{2+} ions and \rm {SO_4}^{2-} ions, while
  • \rm NH_4 OH is made up of \rm {NH_4}^{+} ions and \rm OH^{-} ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

  • \rm Be^{2+} is a cation. In \rm BeSO_4, \rm Be^{2+} was bounded \rm {SO_4}^{2-} anions. During the reaction, it bonds with \rm OH^{-} anions to produce \rm Be(OH)_2.
  • \rm {NH_4}^{+} is also a cation. In \rm NH_4 OH, \rm {NH_4}^{+} was bounded to \rm OH^{-} ions. During the reaction, it bonds with \rm {SO_4}^{2-} anions to produce \rm (NH_4)_2 SO_4.

Hence, the two products will be \rm Be(OH)_2 and \rm (NH_4)_2 SO_4.

Note that charges on the ions must balance. For example, a \rm Be^{2+} ion carries twice as much charge as an \rm {NH_4}^{+} ion. As a result, each \rm Be^{2+} ion would bond with twice as many \rm OH^{-} ions as \rm {NH_4}^{+} would in \rm NH_4 OH.

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Answer:

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luda_lava [24]

Answer:

% differ  1.72%

Explanation:

given data:

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