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Solution
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Mass of the organic compound taken = 0.244 g Mass of CO 2 formed = 0.616 g Mass of H 2 O formed = 0.108 g We know that CO 2 = C and H 2 O = 2H44 g 12 g 18 g 2g (iii) Percentage of O in the compound = 100 - (68.85 + 4.92) = 26.23 The percentage composition of the compoun..
Mass of the organic compound taken = 0.244 g Mass of CO 2 formed = 0.616 g Mass of H 2 O formed = 0.108 g We know that CO 2 = C and H 2 O = 2H44 g 12 g 18 g 2g (iii) Percentage of O in the compound = 100 - (68.85 + 4.92) = 26.23 The percentage composition of the compoun..Solution
Molecular formula = SO 2 Molecular weight = 1 x 32 + 2 x 16 = 64g 64g of SO 2 occupies 22.4L 32 g of SO 2 = ? Volume of 32 g of SO 2 is 11.2 lite..
Molecular formula = SO 2 Molecular weight = 1 x 32 + 2 x 16 = 64g 64g of SO 2 occupies 22.4L 32 g of SO 2 = ? Volume of 32 g of SO 2 is 11.2 lite..Solution
Let the rate of diffusion of oxygen be r(O 2 ) = r 1 The rate of diffusion of the unknown gas r (x) = 4r Molecular mass of O, M(O) = 32 Molecular mass of unknown gas M (x) = M From Graham's Law of diffusion,..
Let the rate of diffusion of oxygen be r(O 2 ) = r 1 The rate of diffusion of the unknown gas r (x) = 4r Molecular mass of O, M(O) = 32 Molecular mass of unknown gas M (x) = M From Graham's Law of diffusion,..Solution
On heating borax with ethanol and concentrated H 2 SO 4 , vapors of triethylborate are produced. These vapors burn with a green edged flame on ignition. The compound triethylborate (C 2 H 5 ) 3 BO 3 is responsible for the green edged flame in the test for borate io..
On heating borax with ethanol and concentrated H 2 SO 4 , vapors of triethylborate are produced. These vapors burn with a green edged flame on ignition. The compound triethylborate (C 2 H 5 ) 3 BO 3 is responsible for the green edged flame in the test for borate io..Solution
So if we calculate the volume of O 2 , which is liberated by 30.36 g of H 2 O 2 at NTP, this will correspond to volume strength of H 2 O 2 . As per the equation: 2 x 34 = 68 g of H 2 O 2 will liberate 22.4 L of O 2 at NTP 68 g of H 2 O 2 will produce O 2 at NTP..
So if we calculate the volume of O 2 , which is liberated by 30.36 g of H 2 O 2 at NTP, this will correspond to volume strength of H 2 O 2 . As per the equation: 2 x 34 = 68 g of H 2 O 2 will liberate 22.4 L of O 2 at NTP 68 g of H 2 O 2 will produce O 2 at NTP..Solution:
The given problem is solved in the following table. Empirical formula = N 2 H 8 S 1 O 4 = N 2 H 8 SO 4..
The given problem is solved in the following table. Empirical formula = N 2 H 8 S 1 O 4 = N 2 H 8 SO 4..Solution:
a. 360 cm 3 of N 2 = 0.45g22.4L of gas = 1 gram molecular weight22.4L = 22,400 cm 3 . (1L = 1000 cm 3 )360 cm 3 of N 2 = 0.45g22,400 cm 3 of N 2 = ? Gram molecular weight of N 2 is 28 g.b. 308 cm 3 Cl 2 = 0.979 g22.400 cm 3 of Cl 2 = ? Molecular weight of Cl 2 = 71...
a. 360 cm 3 of N 2 = 0.45g22.4L of gas = 1 gram molecular weight22.4L = 22,400 cm 3 . (1L = 1000 cm 3 )360 cm 3 of N 2 = 0.45g22,400 cm 3 of N 2 = ? Gram molecular weight of N 2 is 28 g.b. 308 cm 3 Cl 2 = 0.979 g22.400 cm 3 of Cl 2 = ? Molecular weight of Cl 2 = 71...Solution
The frequency ( n ) of emitted light is related to the energy difference of two levels ( D E) as E = 214.68 kJ mol - 1 , h = 39.79 x 10 - 1 4 kJ mol - 1 = 5.39 x 10 1 4 s - 1 8. The wavelength of first spectral line in the Balmer series is 65..
The frequency ( n ) of emitted light is related to the energy difference of two levels ( D E) as E = 214.68 kJ mol - 1 , h = 39.79 x 10 - 1 4 kJ mol - 1 = 5.39 x 10 1 4 s - 1 8. The wavelength of first spectral line in the Balmer series is 65..Solution
Concentration of Zn 2 + ions in solution = 0.1 M Standard electrode potential, E = -0.76 V Reduction reaction: According to Nernst equation, 5. Find the reduction potential of Zn in a solution of 0.01 M ZnSO 4..
Concentration of Zn 2 + ions in solution = 0.1 M Standard electrode potential, E = -0.76 V Reduction reaction: According to Nernst equation, 5. Find the reduction potential of Zn in a solution of 0.01 M ZnSO 4..See what our Users say :
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