Answer:
1.) The nitric acid solution will oxidize and thus dissolve _*(Zn and Pb)*_. This will allow to identify _**Pt**_.
2) To distinguish between _*(Zn and Pb)*_, we can use the nickel nitrate.
3) The nickel nitrate solution will oxidize and thus dissolve _**Zn**_ and will not oxidize or dissolve _**Pb**_.
Explanation:
1) Unlike Zinc and Lead, Platinum does not react with Nitric acid. So, it will be the only metal from step 1 that doesn't react. Pt is identified in this manner.
2) Nickel is higher than Lead in the activity series, but Zinc is higher than both of them in the activity series. This selectivity can be used to distinguish between Zinc and Lead metal powders.
3) Because Zinc is higher than Nickel in the activity series, it means that Zinc metal can and will displace Nickel from Nickel Nitrate solution. Therefore the Nickel Nitrate solution will oxidize and dissolve the Zinc metal.
But, there will be no reaction with the Lead metal powders sample as Pb is lower than Ni in the activity series, so, Nickel Nitrate solution will not oxidize or dissolve the Lead metal powders.
Answer:
ceasar said is a homogeneous mixture
Answer:
the grams of H2O is 351g are processed form 9.75moles of H2
Answer:
Alternative C would be the correct choice.
Explanation:
- The dual compounds were evaluated on something like a TLC plate through three separate additives in conducting a TLC study of ferrocene versus acetylferrocene.
- The polar as well as nonpolar ferrocene where nonpolar is about 0.63 with the maximum
value, and indeed the polar is somewhere around 0.19 with . - TLC plate (30:1 toluene/ethanol) established with.
The other three choices are not related to the given circumstances. So that option C would be the appropriate choice.
