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zmey [24]
3 years ago
14

How much silver can be produced from 125g of Ag2S

Chemistry
2 answers:
vladimir2022 [97]3 years ago
5 0

Answer:

First, you know that for every one Sulfide in the compound there are two silvers.  Now, you need to know the molecular weight of this compound which is 248 g/mol.  (The weight of silver is 108g.  multiply this by two to get the total weight of Ag in Ag2S, which is 216g, and add this to the weight of sulfur, which is 32, and you get 248g/mol).  Now you have to set up a proportion.  

There are 216g of silver in 248g of Ag2S, how many grams of silver are there in 125g of Ag2S?

216g Ag / 248g Ag2S = x / 125 g Ag2S

x = 109g Ag.

Explanation:

rosijanka [135]3 years ago
3 0

Answer:

108.9g of Silver can be produced from 125g of Ag2S

Explanation:

The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1

•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.

•Molar mass of Silver Ag= 107.9g/mol

•Molar mass of Sulphur S= 32g/mol

•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol

•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol

•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S

If 247.8g Ag2S produced 215.8g Ag

125g Ag2S will produce xg Ag

cross multiplying we have

xg= 215.8g × 125g / 247.8g

xg= 26975g/247.8

xg= 108.85g

Therefore, 108.9g of Silver can be produced from 125g of Ag2S

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If the initial [NO2] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M. If the initial is 0.260 , it
Nitella [24]

The question is incomplete, here is the complete question:

At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas

NO_2\rightarrow NO+\frac{1}{2}O_2

The reaction is second order for NO_2 with a rate constant of 0.543M^{-1}s^{-1} at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M

a) 1.01    b) 5.19     c) 0.299      d) 0.0880     e) 3.34

<u>Answer:</u> The time taken is 5.19 seconds

<u>Explanation:</u>

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.543M^{-1}s^{-1}

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.150 M

[A]_o = Initial concentration = 0.260 M

Putting values in above equation, we get:

0.543=\frac{1}{t}\left (\frac{1}{(0.150)}-\frac{1}{(0.260)}\right)\\\\t=5.19s

Hence, the time taken is 5.19 seconds

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