Class 11 and 12

Word Problem based question

This question is word problem of organic chemistry. The question is provided below: Each of the compounds below is treated separately with excess NaBH4. The product of each reaction is then heated with excess concentrated H2SO4. In each case, one or more products are formed with molecular formula C7H10. Which compounds give only one final product with the molecular […]

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Law of multiple proportions

Law of multiple proportions states that if two elements form more than one compound, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers. This law is sometimes called Dalton’s Law, named after John Dalton, the chemist who first expressed it. 2.66 g chloride

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Calculate the number of sulphate ions in 100 ml of 0.001 M H2SO4 solution.

We are given with,   Molarity = 0.001 M   Volume = 100 ml = 0.1 L   Number of moles of H2SO4 =  Molarity*Volume = 0.001 M*0.1 L = 0.0001 mole   Dissociation of sulphuric acid occurs as: H2SO4 ⟶ 2H+ + SO42-   From the above equation, 1 molecule H2SO4 given 1 SO42- ion   then,

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From 200 mg of CO2, 10^21 molecules are removed. How many moles of CO2 are left?

Given mass of CO2 = 200 mg = 0.2 g   Molar mass of CO2 = 44 g/mol   Number of moles of 200 mg CO2 = Given mass/Molar mass of CO2 = 0.2 g/44 g mol-1 = 0.00454 mole   Number of molecules of CO2 removed = 1021   Number of moles of CO2 removed = 1021/(6.022*1023) =

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Assertion-Reason based electronic configuration

Q. Assertion : The element having with electronic configuration 1s2 2s2 2p6 3s2 and 1s2 2s2 belong to the same group    Reason: Both have same number of electrons in the valence shell   Solution:    Elements having same number of valence shell are present in same group.   Here, each elements have 2 valence electrons.

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30% (w/v) of H2SO4 solution. Calculate % (v/w), if density of H2SO4 is 1.05 g/ml and density of solution is 1.5g/ml.

30% w/v H2SO4 means 30 g H2SO4 in 100 ml solution.   Mass of H2SO4 = 30 g Density of H2SO4 = 1.05 g/ml Volume of H2SO4 = Mass/Density = (30 g)/(1.05g/ml) = 28.57 ml   Volume of solution = 100 ml Density of solution = 1.5 g/ml Mass of solution = Volume*Density = (100 ml)*(1.5 g/ml) = 150

30% (w/v) of H2SO4 solution. Calculate % (v/w), if density of H2SO4 is 1.05 g/ml and density of solution is 1.5g/ml. Read More »

Why KMnO4 act as a good oxidizing agent in acidic medium?

In KMnO4, oxidation state of Mn is +7. As the oxidation state of atom increases, the element become more electronegative in nature. Mn gain 5 electron in acidic medium and reduces to +2 oxidation state.   Gain of 5 electron shows that the electronegativity of Mn has increased. Therefore, KMnO4 is a good oxidizing agent.   Note:

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E° for copper electrode is +0.34 V, how will you calculate its emf value?

Question: If E° for copper electrode is +0.34 V, how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu2+ ions in the solution is decreased? Solution: The emf of an electrode when dipped in

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How can you determine limiting molar conductivity for strong and weak electrolyte?

Molar conductivity increases with increase in volume and when volume increases the concentration of solution decreases (because no.of ions in the solution decrease) which makes the solution dilute.   On dilution when concentration approaches zero, the molar conductivity is known as limiting molar conductivity.   In case of strong electrolytes the ionisation is easy and

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