From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
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:
2266g
Explanation:
mass = no.of molecules /6.o23*1o(23) * molar mass
molar mass of co2= 44g /mol
1.5 .10^25/6.023 .10^23 =51.5 moles of co2
51.5 .44g/mol =2266 g
Answer:
We need 17.2 L of Ca(OH)2
Explanation:
Step 1: Data given
Concentration of Ca(OH)2 = 1.45 M
Moles of H2SO4 = 25.0 moles
Step 2: The balanced equation
Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4
Step 3: Calculate moles Ca(OH)2
For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4
For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4
Step 4: Calculate volume of Ca(OH)2
Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2
Volume Ca(OH)2 = 25.0 moles / 1.45 M
Volume Ca(OH)2 = 17.2 L
We need 17.2 L of Ca(OH)2
Answer:
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