The rate of effusion of a particular gas whas measured to be 40ml/min. Under same condition the rate of effusion of pure methane was 20ml/min. Then find molar mass of gas.
A. 4amu
B. 2amu
C. 4gm
D. 2gm
Correct Answer - B `(k_(T_(2)))/(k_(T_(1)))=(mu)^(DeltaT//10)` `(10^(-3))/(k_(T_(1)))=(mu)^(10//10)=2` `k_(T_(1))=(10^(-3))/(2)=0.5xx10^(-3)=5xx10^(-4)"min"^(-1)`
2 Answers 1 viewsCorrect Answer - B Remainin concetration of reactant after 20 min `=(1)/(4)xx102.5"mol"L^(-1)` `"Rate"=kxx["Reactant"]=(0.693)/(t_(1//2))xx["Reactant"]` `=(0.693)/(10xx2.5=0.0693xx2.5 "mol"L^(-1) min^(-1)`
2 Answers 1 views`{:(,NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),("Millimole added",20xx0.2,50xx0.2,,),(,=4,=10,0,0),("Millimole after reaction",0,6,4,4):}` `{"Molarity"}=("millimole")/("Total volume")` `[CH_(2)COOH]=(6)/(70)` &`[CH_(2)COONa]=(4)/(70)` rArr Buffer solution consisting of a weak acid & its salt with a strong base. (2) Let `V mL` of `0.2M NaOH`...
2 Answers 1 views`H_(2)C_(2)O_(4)overset(Delta)toCO+CO_(2)+H_(2)O` `CO+Cl_(2)toCOCl_(2)overset(NH_(3).Delta)to NH_(2)CONH_(2) overset(NH_(3).Delta)larr CO_(2)`
2 Answers 1 viewsEnergy consumption of 50 families per day `=50xx20,000kJ=1xx10^(6)kJ` 809 kJ of energy is obtained by burning methane=16g `1xx10^(6)kJ` of energy will be obtained by burning methane `=(16)/(809)xx10^(6)=1.98xx10^(4)g` `=19.8kg` Since, methane...
2 Answers 1 viewsx litre `toCH_(4)," mole of "CH_(4)=x//22.4` `(10-x)litre to C_(2)H_(6)," mole of "C_(2)H_(6)=(10-x)//22.4` Heat evolved `=(x)/(22.4)xx894+((10-x))/(22.4)xx1500` `474.6=(x)/(22.4)xx894+((10-x))/(22.4)xx1500` `x=0.745,%CH_(4)=74.5%`
2 Answers 1 viewsCorrect Answer - A Combustion of `CH_(4) and C_(2)H_(4)` may be represented as. `"Combustion of "CH_(4) "may be represented as",` `underset(xmL)(CH_(4)(g))+2O_(2)(g) to underset(xmL)(CO_(2)(g))+2H_(2)O(g)` `underset(ymL)(C_(2)H_(4)(g))+3O_(2)(g) to underset(ymL)(2CO_(2)(g))+2H_(2)O(g)` Given, x+y=30 x+2y=40 `therefore x=20,...
2 Answers 1 viewsCorrect Answer - A `underset(1 "L")underset(1"vol")(C_(3)H_(8))(g) + und
2 Answers 1 views`(r_(o_(2)))/(r_(CH_(4)))=(n_(o_(2)))/(n_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(O_(2))))` `=(3)/((2)/(2))=sqrt((16)/(32))` `=(3)/(4sqrt(2))`
2 Answers 1 viewsPM=dRT `(PM)/(d)=` constant (for constant T) `(P_(A)M_(A))/(d_(A))=(P_(B)M_(B))/(d_(B))` `(P_(A))/(P_(B))=(d_(A))/(d_(B))xx(M_(B))/(M_(A))=3xx2=6`
2 Answers 1 views