20 gram of an acid furnishes 0.5 mole of H3O+ ion in its aqueous solution, the value of 1 equivalent of the acid will be 2. when a metal is burnt its weight is increased by 24%, the equivalent weight of the metal will be... 3. 0.7 gram of Na2CO3.xH20 and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCL for complete neutralization, the value of x is... 4. 1 mole of Potassium chlorate is thermally decomposed and excess of Aluminium is burnt in the gaseous product ,how many mole of Aluminium oxide are formed? 5. The molality of 1 litre solutions with X% percent by weight H2SO4 is equal to 9. the weight of the solvent present in the solution is 910 grams, the value of x is6. 1 litre of CO2 is passed over hot coke. the Volume becomes 1.4 litre, the percentage composition of products is I.e. of CO2 and CO (in litres)7. The weight of 350 ml of a diatomic gas at 0 degree Celsius and 2 ATM pressure is 1 gram, the weight of one atom is....a. 16/NA b. 32/NA c. 16NA d. 32NA( Where NA is Avogadro`s constant)8. 0.078g aluminium Hydroxide is dehydrated to Al2O3. Al2O3 so obtained reacted with 6 mili equivalent of HCL, the equivalent of AlCl3 produced during the reaction-9. Amount of oxygen in 32.2 gram of Na2SO4.10 H2O is....10. A gaseous alkane was exploded with oxygen the volume of O2 for complete combustion of CO2 formed was in the ratio 7: 4 the molecular formula of Alkane is,11. Vapour density of metal chloride is 66, its oxide contains 53 % metal the atomic weight of metal is its oxide contains 53 % metal, the atomic weight of metal is...12. 0.05 moles of Na2CO3 will react with how many equivalent of magnesium hydroxide....13. How many millilitres of 0.5 molar H2SO4 are needed to dissolve 0.5 gram of copper (II) carbonate?14. V 1 ml of sodium hydroxide of normality X and V 2 ml of Barium Hydroxide of normality Y are mixed together, the mixture is completely neutralized by hundred ml of 0.1 normal HCL. If V1/V2=1/4 and X/Y = 4 what fraction of acid is neutralized by barium hydroxide.15.The sea water contains 63 ppm NO3- ions and its density is 1.01 gram/ml then approximate molarity of sea water is...16. 3g of An oxide of a metal is converted completely to 5g chloride, then the equivalent weight of metal is.....

Let's tackle these chemistry problems step by step. Each question involves different concepts, so I'll break them down clearly for you.

1. Equivalent Weight of the Acid

Given that 20 grams of an acid furnishes 0.5 moles of H₃O⁺ ions, we can determine the equivalent weight of the acid. The equivalent weight is defined as the weight of the acid that furnishes one mole of H₃O⁺.

Since 0.5 moles of H₃O⁺ corresponds to 20 grams of acid, we can find the equivalent weight:

• 1 equivalent = 2 moles of H₃O⁺ (as given).
• Thus, 20 grams corresponds to 0.5 moles, meaning 1 mole would be 40 grams.
• Therefore, the equivalent weight = 40 grams / 2 = 20 grams.

2. Equivalent Weight of the Metal

When a metal is burnt and its weight increases by 24%, we can find the equivalent weight using the formula:

• Let the initial weight of the metal be W.
• After burning, the weight becomes W + 0.24W = 1.24W.
• The increase in weight (0.24W) corresponds to the weight of oxygen that combined with the metal.

To find the equivalent weight, we can use the formula:

Equivalent Weight = Weight of Metal / Number of Equivalents

Assuming the metal reacts with one equivalent of oxygen, the equivalent weight will be:

Equivalent Weight = W / 1 = W

Since we don’t have the actual weight of the metal, we can’t calculate a numerical value without additional data.

3. Value of x in Na₂CO₃.xH₂O

To find the value of x, we start with the neutralization reaction. We know that 20 ml of the Na₂CO₃.xH₂O solution required 19.8 ml of N/10 HCl for complete neutralization.

First, calculate the number of equivalents of HCl used:

• Normality of HCl = 0.1 N.
• Volume of HCl = 19.8 ml = 0.0198 L.
• Equivalents of HCl = Normality × Volume = 0.1 × 0.0198 = 0.00198 equivalents.

Since Na₂CO₃ provides 1 equivalent per mole, the equivalents of Na₂CO₃ in 20 ml is also 0.00198. Therefore, the molarity of the Na₂CO₃ solution is:

Molarity = Equivalents / Volume = 0.00198 / 0.020 = 0.099 M.

Now, we can relate this to the weight of Na₂CO₃.xH₂O:

Let the molar mass of Na₂CO₃ be approximately 106 g/mol. The weight of Na₂CO₃.xH₂O in 100 ml is:

Weight = Molarity × Molar Mass × Volume = 0.099 × 106 × 0.1 = 1.05 g.

Now, if we assume the molar mass of water is 18 g/mol, we can set up the equation:

1.05 = 106 + 18x.

Solving for x gives:

x = (1.05 - 106) / 18 = 0.05.

Thus, x is approximately 0.

4. Moles of Aluminium Oxide from Potassium Chlorate Decomposition

When 1 mole of potassium chlorate (KClO₃) decomposes, it produces potassium chloride (KCl) and oxygen (O₂):

2 KClO₃ → 2 KCl + 3 O₂

From this reaction, we see that 2 moles of KClO₃ yield 3 moles of O₂. If we burn excess aluminum in the gaseous product, the reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

From the stoichiometry, 3 moles of O₂ yield 2 moles of Al₂O₃. Therefore, if we have 1 mole of KClO₃, we can find the moles of Al₂O₃ produced:

• 1 mole of KClO₃ produces 1.5 moles of O₂ (from 2 KClO₃ gives 3 O₂).
• 1.5 moles of O₂ will produce 1 mole of Al₂O₃ (using the 4:3 ratio).

Thus, 1 mole of Al₂O₃ is formed from the decomposition of 1 mole of KClO₃.

5. Finding