White precipitate of `PbSO_(4)` gets dissolved in:
A. concentrated `H_(2)SO_(4)` on heating
B. concentrated `NaOH`
C. `(NH_(4))_(2)CO_(3)`
D. Dilute `HNO_(3)`
1 Answers
Talk Doctor Online in Bissoy App
A substance X containing three carbon atoms gives white crystalline precipitate with sodium sulphite but does not give red precipitate with Fehling so
Correct Answer - B `underset((X))(CH_(3)COCH_(3))overset(NH_(2)-NH_(2))underset(NaOH)rarrunderset("Propane")(CH_(3)CH_(2)CH_(3)+H_(2)O)`
2 Answers 1 views
A white salt gives a white precipitate with barium chloride solution which is insoluble in ammonia and hot water. The salt is
Correct Answer - B `BaCl_2` with `MgSO_4` frroms `BaSO_4` which is insoluble in water due to its small hydration energy.
2 Answers 1 views
A metal (M) burns with dazzling brilliance in air to give a white powder.The white powder reacts with water to form a white precipitate and a colourle
Correct Answer - C Hydrogen is liberated when Mg reacts with hot water. Rb, K and Ca gives hydrogen gas even with cold water. `Mg+O_2toMgO , Mg+N_2toMg_3N_2` `Mg_3N_2+6H_2O to 3Mg(OH)_2 +...
2 Answers 1 views
A metal nitrate solution reacts with dilute hydrochloric acid to give a white precipitate which is soluble in concentrated potassium chloride.White pr
Correct Answer - A `Pb^(2+)+2Cl^(-)toPbCl_2darr ("white"), PbCl_2+ 2Cl^(-)to[PbCl_4]^(2-)` (colourless soluble complex ) `PbCl_2darr +H_2StoPbSdarr`(black)+2HCl `PbS+4H_2O_2toPbSO_4`(white)+`4H_2O` `PbSO_4darr +2CH_3COONH_4to(NH_4)_2SO_4+(CH_3COO)_2Pb`
2 Answers 1 views
Three bags contain a number of red and white balls as follows Bag I : 3 red balls, Bag II : 2 red balls and 1 white balls and Bag III : 3 white balls.
Bag I : 3 red balls and 0 white ball Bag II : 2 red balls and 1 white ball. Bag III : 0 red ball andf 3 white balls....
2 Answers 2 views
There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal pr
Let `U_(1)`={2 white, 3 black balls} `U_(2)`={3 white , 2 black balls} and `U_(3)`={4 white ,1 black balls} `thereforeP(U_(1))=P(U_(2))=P(U_(3))=1/3` Let `E_(1)` be the event that a ball is chosen from...
2 Answers 1 views
A white precipitate (X) is formed when a mineral (A) is boiled with `Na_(2)CO_(3)` solution. The precipitate is filtered and filtrate contains two com
Correct Answer - A::B::C::D The reactons are : `underset((A))(Ca_(2)B_(4)O_(11))+2Na_(2)CO_(3)(aq)tounderset((B))(2CaCO_(3))darr+underset(( C))(Na_(2)B_(4)O_(7))+underset((D))(2NaBO_(2))` `underset((D))(4NaBO_(2))+CO_(2) to underset(( C))(Na_(2)B_(4)O_(7))+Na_(2)CO_(3) , underset((C ))(Na_(2)B_(4)O_(7))overset(Delta) to underset((D))(2NaBO_(2))+underset((E))(B_(2)O_(3))` `underset((E))(B_(2)O_(3))+CoO to underset("Blue")(Co(BO_(2))_(2)(F))`
2 Answers 1 views
A white crystalline substance dissolves in water.On passing `H_(2)S` in this solution, a black precipitate is obtained.The black precipitate dissolves
Correct Answer - C `Pb^(2+)+H_(2)StoPbSdarr("black")+2H^(+), 3PbS+8HNO_(3)to3Pb(NO_(3))_(2)+2NOuarr+3Sdarr+4H_(2)O` `Pb^(2+)+SO_(4)^(2-) to PbSO_(4) darr` (white)
2 Answers 1 views
`MgSO_(4) +NH_(4)OH + Na_(2)HPO_(4) to` white crystalline precipitate.The formula of crystalline precipitate is:
Correct Answer - C `MgSO_(4)` on reaction with `Na_(2)HPO_(4)` in presence of `NH_(4)OH` gives white precipitate of `Mg(NH_(4))PO_(4)` `MgSO_(4)+Na_(2)HPO_(4) + NH_(4)OH to Mg(NH_(4))PO_(4)darr+Na_(2)SO_(4)+H_(2)O`
2 Answers 1 views
Calculate `K_(sp)` if `(PbSO_(4))E_(cell)` at 298 K of this electrode is 0.236 V `Pb(s)|PbSO_(4)(s)|underset(0.01M)(Na_(2)SO_(4)(aq))||underset(0.1M)(
Let it be conc. Cell Anode `Pb(s)toPb_((a))^(2+)+2e^(-)` cothode `Pb_((as))^(2+)+2e^(-)toPb(s)` `Pb_((c))^(2+)hArrPb_((a))^(2+)` `E_(cell)=E_(cell)^(0)-(0.059)/(2)log[((pb^(2+))_(a))/((Pb^(2+))_(c))]` `0.236=(0.059)/(2)log[((Pb^(2+))_(a))/((Pb^(2+))_(c))]` `(Pb^(2+))_(a)=10^(-9)M` `K_(sp)=(Pb^(2+))_(a)(SO_(4)^(2-))_(a)=10^(-11)` Let if it is not a conc cell Anode: `Pb(s)+SO_(4)^(2-)(aq)toPbSO_(4)(s)+2e^(-)` cathode: `Pb^(2+)(aq)+2e^(-)toPb(s)` `Pb^(2+)(c)+SO_(4)^(2-)(a)toPbSO_(4)(s)` `E_(cell)^(0)=E_(Pb^(2+)//Pb)^(0)-E_(SO_(4)^(2-)|PbSO_(4)|Pb)^(0)=E_(Pb^(2+)|Pb)^(0)-[E_(Pb^(2+)|Pb^(+))^(0)+(0.059)/(2)logK_(sp)]` `E_(cell)^(0)=-(0.059)/(2)logK_(sp)` `0.236=(0.059)/(2)logK_(sp)logK_(sp)-(0.059)/(2)log[(1)/((Pb^(2+))_(c)(SO_(4)^(2-))_(a))]=log[(K_(SP))/(0.1xx0.01)]=log[(K_(SP))/(10^(-3))]log10^(-8)`
2 Answers 1 views