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Kaylis [27]
3 years ago
15

What are the masses of the four different BrClBrCl molecules? Express the masses in atomic mass units using six significant figu

res, in decreasing numeric order (highest to lowest), separated by commas.

Chemistry
1 answer:
Sliva [168]3 years ago
6 0

Answer ; The question is missing in some details, but here are he details ;

The two naturally occurring isotopes of bromine are

81Br (80.916 amu, 49.31%) and

79Br (78.918 amu, 50.69%).

The two naturally occurring isotopes of chlorine are

37Cl (36.966 amu, 24.23%) and

35Cl (34.969 amu, 75.77%).

Bromine and chlorine combine to form bromine monochloride, BrCl.

Explanation:

The detaile calculation is as shown in the attachment.

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Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi
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Answer:

                  Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

                                    Copper Metal  =  Cu

                                    Nitric Acid  =  HNO₃

                                    Copper (II) Nitrate  =  Cu(NO₃)₂

                                    Nitrogen Dioxide  =  NO₂

                                    Water  =  H₂O

Step 2: Write down the unbalance Chemical equation,

                         Cu  +  HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

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Answer:

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