A chalk sample exactly 17.52 gram HCl for complete reaction with all `CaCO_(3)` present in it. If the chalk sample is 72% pure, the mass of sample taken is
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Lime is made commercially by decomposition of limestonese, `CaCO_(3)`. What is the change in intremal energy when `1.00` mole of soild `CaCO_(3)` (V=
Correct Answer - `q = 177.9 kJ, w = - 2.5 kJ; DeltaE = 175.4 kJ` `q = 177.9 kJ` `w =- P_("ext") (V_(2)- V_(1)) =- (24.4 -(34.2 + 16.9 )...
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Calculate the molecular mass of a substance, 1.0 gram of which when dissolved in 100 gram of solvent gace an elecation of 0.307 K in the boiling point
Correct Answer - 59.93 g `mol^(-1)` `W_(B)=1.0g, W_(A)=0.1 kg, DeltaT_(b)=0.307 K, K_(b)=1.84 K//m, M_(B)=?` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.84" K kg mol"^(-1))(1.0g))/((0.307 K)xx(0.1 kg))=59.93" g mol"^(-1)`.
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X gram of pure `As_2S_3` is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hot acidified `KMnO_4` then X, mass of `As_2S
Correct Answer - B `5As_2S_3+28KMnO_4+H^(+)to10H_3AsO_4+28Mn^(2+)+SO_4^(2-)` meq of `As_2S_3`=meq. of `KMnO_4` `X/246xx1000xx28=50xx0.1xx5 " " X=0.22 g`
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If 20g of `CaCO_(3)` is treated with 20g of HCl. How many grams of `CO_(2)` can be generated according to the following equations? `CaCO_(3)+2HCl(aq)
`underset(100g)underset("1 mol")(CaCO_(3))+underset(73g)underset("2 mol")(2HCl(aq)) to CaCl_(2)(aq.)+H_(2)O(l)+underset(44g)underset("1 mol")(CO_(2)g)` Let `CaCO_(3)(s)` be completely consumed in the reaction. `therefore 100g CaCO_(3) "give" 44gCO_(2)` `therefore 20g CaCO_(3)"will give" (44)/(100)xx20g CO_(2)=8.8CO_(2)` Let HCl be completely consumed....
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Calculate the mass of oxygen obtained by complete decomposition of 10kg of pure potassium chlorate (Atomic mass K=39,O=16 and Cl=35.5)
Correct Answer - C The reaction is : `underset(2xx122.5 kg)(2KClO_(3))(s) to 2KCl(s) + underset(3xx32 kg)(3O_(2)(g))` Mass of `O_(2)` obtained by 10 kg `KClO_(3) = (96 xx 10)/(245) = 3.92 kg `]
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1 gram of carbonate `(M_(2)CO_(3))` on treatment with excess HCl produces 0.1186 mole of `CO_(2)`. The molar mass of `M_(2)CO_(3) "in g mol"^(-1)`
Correct Answer - B The reaction is : `M_(2)CO_(3)+ 2HCl to 2Mcl + H_(2)O + CO_(2)` Number of moles of `CO_(2)` = Number of moles of `M_(2)CO_(3)` `0.01186 = ("Mass")/("Molar mass...
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How much mass (kg) of `CaCO_(3)` is needed to produce 336 kg of Cao ? [Given : % yield of reaction = 60]
Correct Answer - A `CaCO_(3)(s)rarrunderset(336kg)(CaO(s))+CO_(2)1g` Moles of CaO = `(336)/(56)xx10^(3)=6xx10^(3)` Mass of `CaCO_(3)` required `=6xx10^(3)xx(100)/(60)xx100=10^(3)kg`
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Equation `1+x^2+2x"sin"(cos^(-1)y)=0` is satisfied by exactly one value of `x` exactly two value of `x` exactly one value of `y` exactly two value of
Correct Answer - A::C Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0` Since x is real, `D ge 0`. Therefore, `4(sin(cos^(-1)y))^(2) -4 ge 0` or `(sin(cos^(-1)y))^(2) ge 1`...
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A 20 gm mixture of Ca`(OH)_(2)` and `CaCl_(2)` require 50ml 2M HCl for complete reaction Then what will be the mass % of `Ca(OH)_(2)`
`n_(Cal(OH)_(2))=a n_(CaCl_(2))=b` `74a+111b=20`. . . .(i) `Ca(OH)_(2)+2HClrarrCaCl_(2)+2H_(2)O` `n_(Ca(OH)_(2))=a=(n_(HCl))/(2)=(50xx2)/(1000xx2)=(1)/(20)=(0.1)/(2)` `W_(Ca(OH)_(2))=(7.4)/(2)gm=3.7` Mass `% Ca(OH)_(2))=(3.7)/(20)xx100=18.5%`
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A bag contains 20 coins. If the probability that the bag contains exactly 4 biased coin is 3/4 and that of exactly 5 biased coin is 2/3, then the prob
Correct Answer - B P (4 bases coin)`=1/3` P(5 biased coin) `=1/4` Hence, the required probability is `=1/3(""^(4)C_(3)""^(16)C_(6))/(""^(20)C_(9))+2/3(""^(5)C_(4)""^(15)C_(5))/(""^(20)C_(9))(1)/(""^(11)C_(1))` `=2/33[(""^(16)C_(6)+5""^(15)C_(5))/(""^(20)C_(9))]`
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