Answer:
- Current market value of silver produced in the experiment = $0.2763
- Check Explanation for the appearance of leftover copper wire used in the experiment.
Explanation:
Price of Silver on today's market = $15.34 per troy ounce
Amount of Silver produced in the experiment = 0.56 g
31.1 g = 1 troy ounce
0.56 g = (0.56/31.1) = 0.01801 troy ounce
1 troy ounce = $15.34
0.01801 troy ounce = 0.01801 × $15.34 = $0.2763
The original reaction to extract the silver is one between Copper metal and the Silver Nitrate.
The Copper metal dissolves in the Silver Nitrate to give Copper Nitrate thereby displacing the Silver ion to become silver metal which we just valued above. But copper nitrate is a blue solution, So, it is a good estimate to say that the entire solution changed to that blue colour and the appearance of the leftover copper metal is colour blue. The silver metal forms as fractal-like crystals on this leftover copper.
Hope this Helps!!!
<u>Answer: </u>He sorted on the basis of average atomic weights.
<u>Explanation:</u>
Dmitri Mendeleev in 1863 arranged chemical elements in the groups according to their average atomic weights.
He said that elements which have similar chemical properties either have similar atomic weights or their atomic weights increases regularly.
Hence, he collaborated a great work in sorting the elements according to their average atomic weights.
Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
4 infiltration percolation!! I think! Correct me if I’m wrong
I am going to say it is false.
