Answer:
Option (E) is correct
Explanation:
Solubility equilibrium of 
is given as follows-
Hence, if solubility of is S (M) then-
![[Pb^{2+}]=S(M)](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3DS%28M%29)
and ![[IO_{3}^{-}]=2S(M)](https://tex.z-dn.net/?f=%5BIO_%7B3%7D%5E%7B-%7D%5D%3D2S%28M%29)
Where species under third bracket represent equilibrium concentrations
So, solubility product of , ![K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BIO_%7B3%7D%5E%7B-%7D%5D%5E%7B2%7D)
Here, ![[Pb^{2+}]=S(M)=5.0\times 10^{-5}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3DS%28M%29%3D5.0%5Ctimes%2010%5E%7B-5%7DM)
So, ![[IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M](https://tex.z-dn.net/?f=%5BIO_%7B3%7D%5E%7B-%7D%5D%3D2S%28M%29%3D%282%5Ctimes%205.0%5Ctimes%2010%5E%7B-5%7D%29M%3D1.0%5Ctimes%2010%5E%7B-4%7DM)
So, 
Hence option (E) is correct
Answer:
N2I6 = 789 g
N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
N2I6=0.136molesx789g/mole=107g=110g
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Hope this helps, have a BLESSED AND WONDERFUL DAY!
- Cutiepatutie ☺❀❤
Answer:
first of all is that really chemistry
Answer:
B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-
Explanation:
In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.
the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.
Answer:
Chloroform= limiting reactant
0.209mol of CCl4 is formed
And 32.186g of CCl4 is formed
Explanation:
The equation of reaction
CHCl3 + Cl2= CCl4 + HCl
From the equation 1 mol of
CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4
From the question
25g of CHCl3 really with Cl2
Molar mass of CHCl3= 119.5
Molar mass of Cl2 = 71
Hence moles of CHCl3= 25/119.5 = 0.209mol
Moles of Cl2 = 25/71 = 0.352mol
Hence CHCl3 is the limiting reactant
Since 1 mole of CHCl3 gave 1mol of CCl4
It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4
Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g
