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Scorpion4ik [409]
2 years ago
9

How many moles of ammonium ions are in 6.985 g of ammonium carbonate?

Chemistry
1 answer:
Mila [183]2 years ago
5 0
 <span>(NH4)2CO3 -> 96.09 g/mol 

(6.995g ammonium carbonate)(1mol ammonium carbonate/ 96.09 g ammonium carbonate) = 0.072796 mol ammonium carbonate 

In this conversion, the 'grams' unit is crossed out because it is in both the numerator and the denominator, which leaves the 'mol' unit left. 

Looking at the formula (NH4)2CO3, you can look at it as if it were: 
2 mol (NH4) + 1 mol (CO3) = 1 mol (NH4)2CO3 

For every 1 mol of ammonium carbonate, you have 1 mol of carbonate ions and 2 moles of ammonium ions. 

(0.072796 mol ammonium carbonate) = (0.072796 mol carbonate ion)+(0.363981 mol ammonium ion) </span>
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The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
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Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

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123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

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The final step is adding the value to get the average atomic weight.

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3 years ago
Consider the following data for five hypothetical elements: Q, W, X, Y, and Z. Rank the elements from most reactive to least rea
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Y, Q, W, Z and X

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Q + w+ Reaction occurs

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Therefore, the most reactive to the least reactive is arrange as follows.

Y, Q, W, Z and X

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