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gulaghasi [49]
3 years ago
8

How many grams of silver nitrate are needed to react with 156.2g of sodium sulfide to produce 595.8g of silver sulfide and 340.0

g of sodium nitrate
Would really appreciate it if you can answer this question and explain it for me, thankkks
Chemistry
1 answer:
qaws [65]3 years ago
3 0
Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃

Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate

Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams
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Read 2 more answers
The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances
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Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:  

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu  

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 +  X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

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3 years ago
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