📖 Samacheer Kalvi · 11th TN - English Medium · Chemistry Volume 1 · Page 246question

Unit - 1 Basic Concepts of Chemistry and Chemical Calculations · Part 5

Chapter 3: 7 · Chemistry Volume 1

CH COOH → CH COONa + H O + CO ↑ . g g → g + x The amount of CO released, x = . g No. of moles of CO released = .

/ = . mol H (g) + Cl (g) → HCl (g) Content H (g) Cl (g) HCl (g) Stoichiometric coefficient No. of moles of reactants allowed to react at K and atm pressure . L ( mol) .

L ( . mol) - No. of moles of reactant reacted and product formed . .

Amount of HCl formed = mol 11th Std Chemistry 11th Std Chemistry - - - - . BaCl + H S O Ba S O + HCl → . P + NaOH + H O H + NaH PO → P . The reduction reaction of the oxidising agent(MnO – ) involves gain of electrons.

Hence the equivalent mass = (Molar mass of KMnO )/ = . / = . . No.

of moles of water present in g = Mass of water / Molar mass of water = g / g mol - = moles One mole of water contains = . x water molecules mole of water contains = . x x = . x water molecules .

. g of gas occupies a volume of . liters at K and atm pressure Therefore, the mass of gas that occupies a volume of . liters ( g L L g Molar mass of NO + ) = g .

No. of electrons present in one ammonia (NH ) molecule ( + ) = No of molesof ammonia Molar mass gmol Mass g . mol No of molecules present in .1mol of ammonia = . .

= . No of electrons

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