Solution
Volume of 1.6 mole of water at 373 K in gaseous state Volume of 1 mol = 18 g of liquid water (density = 1 g ml - 1 ) = 18 x 1.6 x 10 - 3 L = 0.0288 L Now work done (W) = -P(V 2 - V 1 ) = -1(48.93 - 0.0288) = - 48.90 atm L = - 48.90 x 101.325 J..
Volume of 1.6 mole of water at 373 K in gaseous state Volume of 1 mol = 18 g of liquid water (density = 1 g ml - 1 ) = 18 x 1.6 x 10 - 3 L = 0.0288 L Now work done (W) = -P(V 2 - V 1 ) = -1(48.93 - 0.0288) = - 48.90 atm L = - 48.90 x 101.325 J..Solution:
2 vol. : 1vol : 2 vol. 30 litres: x litres : y litres CO : O 2 CO : CO 2 2 : 1 2 : 2 30 : x 30 : y 2x = 30 x 1 2y = 30 x 2 x = 15 y= 30 Volume of oxygen = 15 litres Volume of carbon dioxide =..
2 vol. : 1vol : 2 vol. 30 litres: x litres : y litres CO : O 2 CO : CO 2 2 : 1 2 : 2 30 : x 30 : y 2x = 30 x 1 2y = 30 x 2 x = 15 y= 30 Volume of oxygen = 15 litres Volume of carbon dioxide =..Solution:
If we replace y by , we have y'+ y \ f (x) satisfies the equation y' + y = 2 Moreover, ..
If we replace y by , we have y'+ y \ f (x) satisfies the equation y' + y = 2 Moreover, ..Solution
Here the products are X and Y and reactants are A and B. The powers of the concentrations of A, B, X and Y correspond to the respective number of moles in the equation. 6. Write down the expression for the equilibrium constants What i..
Here the products are X and Y and reactants are A and B. The powers of the concentrations of A, B, X and Y correspond to the respective number of moles in the equation. 6. Write down the expression for the equilibrium constants What i..Solution
(iii) The compound, is acid anhydride, ((RCO) 2 O). The parent acid of this compound is CH 3 COOH (ethanoic acid). Therefore, the given compound is named as, Ethanoic anhydride. (iv) The compound, The longest chain in this compound contains four carbon atoms. Therefore, it is named after ..
(iii) The compound, is acid anhydride, ((RCO) 2 O). The parent acid of this compound is CH 3 COOH (ethanoic acid). Therefore, the given compound is named as, Ethanoic anhydride. (iv) The compound, The longest chain in this compound contains four carbon atoms. Therefore, it is named after ..Solution
Gram molecular mass of ammonia (NH 3 ) = (1 x 14) + (3 + 1) = 14 + 3 = 17 g Mass of 1 mol of ammonia = 17g Molar volume = 22.4 litres Volume of 3.4 g of ammonia at STP = ? Mass : Volume 17 g : 22.4 litre 3.4 g : x Volume occupied by 3.h..
Gram molecular mass of ammonia (NH 3 ) = (1 x 14) + (3 + 1) = 14 + 3 = 17 g Mass of 1 mol of ammonia = 17g Molar volume = 22.4 litres Volume of 3.4 g of ammonia at STP = ? Mass : Volume 17 g : 22.4 litre 3.4 g : x Volume occupied by 3.h..Solution:
Formula of hydrous copper sulphate = CuSO 4 .5H 2 O Gram Molecular Mass of CuSO 4 .5H 2 O = (1 x 64) + (1 x 32) + (4 x 16) + 5 {(1 x 2)+ 16} = 64 + 32 + 64 + 90 = 250 g Mass of 1 mole of CuSO 4 .5H 2 O = 250 g No. of m..
Solution
Ag 2 CrO 4 dissolves in water in accordance with the equilibrium. If S is the solubility of Ag 2 CrO 4 then in a saturated solution, Then, K s p (Ag 2 CrO 4 ) = [Ag + ] 2 [CrO 4 2 - ] = (2S) 2 (S) 1.7 x 10 - 1 1 (mol / L) 3..
Ag 2 CrO 4 dissolves in water in accordance with the equilibrium. If S is the solubility of Ag 2 CrO 4 then in a saturated solution, Then, K s p (Ag 2 CrO 4 ) = [Ag + ] 2 [CrO 4 2 - ] = (2S) 2 (S) 1.7 x 10 - 1 1 (mol / L) 3..Solution
The overall energy changes for the formation of MgCl and MgCl 2 in water can be understood by considering the various steps as: H h y d r a t i o n = -735 kJ mol - 1 Net Energy change = - 46 kJ mol - 1 ..
The overall energy changes for the formation of MgCl and MgCl 2 in water can be understood by considering the various steps as: H h y d r a t i o n = -735 kJ mol - 1 Net Energy change = - 46 kJ mol - 1 ..Solution
Therefore simple ratio = Fe 3 O 4 . Empirical formula = Fe 3 O 4..
Therefore simple ratio = Fe 3 O 4 . Empirical formula = Fe 3 O 4..See what our Users say :
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