Some Basic Conc...

H₂O₂ acts as both oxidising as well as reducing agent its product is H₂O, when act as oxidising agent and its product is 0₂ when act as of reducing agent. The strength of '10P means one liter of H₂O₂ solution on decomposition at S.T.P. condition liberate 10 liters of oxygen gas (H₂O → H₂O + 1/2 O₂) 15 gm Mn0₄)₂ sample containing inert impurity is completely reacting with 100 ml of '11.2 V' H₂O₂, then what will be the % purity of Ba (Mn0₄)₂ in the sample?            (Atomic mass Ba = 137, Mn = 55)           

1.2 gm of carbon is burnt completely in oxygen (limited supply) to produce CO and CO₂. This mixture of gases is treated with solid I₂O₅ (to know the amount of CO produced), the liberated iodine required 120 ml of 0.1 M hypo solution for complete titration. The % of carbon converted into CO is:

Nitrogen (N), phosphorus (P) and Potassium (K) are the main nutrients in plant fertilizers. According to an industry convention, the numbers on the label refer to the mass % of N, P₂O₅ and K₂O in that order calculate the N : P : K ratio of a 30 : 10 : 10, fertilizer in terms of moles of each element and express it as x: y : 1.                       

In what ratio should a 15% solution of acetic acid be mixed with a 3% solution of the acid to prepare a 10% solution(all percentages are mass/mass percentages):           

105 ml of pure water at 4°C saturated with NH₃ gas yielded a solution of density 0.9 g/ml and containing 30% NH by mass. Find the volume of resulting NH₃ solution.     

The equivalent mass of H₃BO₃ (M = Molar mass of H₃BO₃) in its reaction with NaOH to from Na₂B₄O₇ is equal to –                       

X gram of pure As₂S₃ is completely oxidised to respective highest oxidation states by 50 ml of 0.1 M hot acidified KMnO₄ then x mass of As₂S₃ taken is : (Molar mass of As₂S₃ = 246)

The number of moles of ferrous oxalate oxidised by one mole of KMnO₄ is

How many moles of KMnO₄ are needed a mixture of 1 mole of each FeSO₄ & FeC₂O₄ in acidic medium

In the reaction  

Na₂S₂O₃ + 4C1₂ + 5H₂O → Na₂SO₄ + H₂S0₄ + 8HCI

the equivalent weight of Na₂ S₂ O₃ will be

(M= molecular weight of Na₂S₂O₃)

C₂H₄ + xO₂ -> 2CO₂ + yH₂O, what is the value of x + y?

The following equations are balanced atomwise and chargewise.

(i) Cr₂O₇^2- + 8H+ + 2H₂O₂ → 2Cr³+ + 7H₂O + 2O₂

(ii) Cr₂O₇^2- + 8H+ + 5H₂O₂→ 2Cr³⁺ + 9H₂O + 4O₂

(iii) Cr₂O₇^2- + 8H+ + 7H₂O₂→ 2Cr+ + 11H₂O + 5O₂

The precise equationl equations representing the oxidation of H₂O₂ is/are

An excess of NaOH was added to 100 mL of a ferric chloride solution. This caused the precipitation of 1,425 g of Fe(OH)₃. Calculate the normality of the ferric chloride solution

In the reaction CrO₅ + H₂SO₄ →Cr₂(SO₄)₃ + H₂O + O₂ one mole of CrO₅ will liberate how many moles of O₂

0.4g of a polybasic acid H_nA (all the hydrogens are acidic) requires 0.5g of NaOH for complete neutralization. The number of replaceable hydrogen atoms in the acid and the molecular weight of 'A' would be : (Molecular weight of the acid is 96 gms.)

A solution of Na₂S₂O₃ is standardized iodimetrically against 0.1262 g of Mr% This process requires 45 mL of the Na₂ S₂0₃ solution. What is the strength of the Na₂S₂O₃?

25.0 g of FeSO₄.7H₂O was dissolved in water containing dilute H₂SO₄, and the volume was made up to 1.0 L. 25.0 mL of this solution required 20 mL of an N/10 KMnO₄ solution for complete oxidation. The percentage of FeSO₄^. 7H₂O in the acid solution is

1.0 mol of Fe reacts completely with 0.65 mol of O₂ to give a mixture of only Fe0 and Fe₂O₃. The mole ratio of ferrous oxide to ferric oxide is

25 mL of a solution containing HCf and H₂S0₄ required 10 mL of a 1 N NaOH solution for neutralization. 20 mL of the same acid mixture on being treated with an excess of AgNO₃ gives 0.1425 g of AgCf. The normality of the HC and the normality of the H₂S0₄ are respectively

If 10 gm of V₂O₅ is dissolved in acid and is reduced to V²⁺ by zinc metal, how many mole of I₂ could be reduced by the resulting solution if it is further oxidised to VO²⁺ ions ?

[Assume no change in state of Zn²+ions] (V = 51, 0 = 16, I = 127) :

0.70 g of mixture (NH₄)₂ SO₄ was boiled with 100 mL of 0.2 N NaOH solution till all the NH₃(g) evolved and get dissolved in solution itself. The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H₂SO₄ solution. The percentage purity of the (NH₄)₂ SO sample is

A mixed solution of potassium hydroxide and sodium carbonate required 15 mL. of an N/20 HCI solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is

The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to, an excess of potassium iodide solution liberating iodine according to the following equation

 2Cu² (aq) + 4I^– (aq)   2CuI(s) + I₂(aq)

The iodine liberated required 24.5cm³ of a 0.100 mole dm^-3 solution of sodium thiosulphate

2S₂0₃^2- (aq) + I₂(aq) → 2I^– (aq) + S₄0₆^2- (aq)

the percentage of copper, by mass in the copper (ll) salt is. [Atomic mass of copper = 63.5]

KIO₃ reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo solution, The reactions are

1. IO₃^– + I^– →I₂                (valency factor = 5/3)

2. I₂ + S₂O₃^2– → S₄O₆^2– + I^–        (valency factor = 2)

meq of hypo = meq of I₂ = meq of IO₃^– + meq of I^–

IO₃^– react with I^– ⇒ meq of IO₃^– = meq of I^–

Statement-1 : meq of hypo = 2 x meq of IO₃^–

Statement-2 : valency factor of I₂ in both the equation are different therefore we cannot equate milliequivalents in sequence

In the reaction:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
how many moles of CO₂ are formed when 1 mole of O₂ is consumed?