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Solution
The fifth period begins with the filling of '5s' orbitals and continues till the filling of sixth energy level (6s) starts. The subshells which follow '5s' are, 4d, 5p, 6s, Thus, the elements which involve filling of 5s, 4d, 5p, sub shells are accommodated ..
Solution:
22.4L of a gas = 1 GMM100 cm 3 of gas = 0.48 g22.4 x 1000 of gas = ? Gram molecular weight of the gas is 107.52..
22.4L of a gas = 1 GMM100 cm 3 of gas = 0.48 g22.4 x 1000 of gas = ? Gram molecular weight of the gas is 107.52..Solution
The second electron in the case of the alkali metals (Na) is to be removed from a cation, which has already acquired a noble gas configuration. On the other hand, in the alkaline Earth metal (Mg), the second electron is to be removed from a monovalent cation (Mg + ), which still has one electron..
The second electron in the case of the alkali metals (Na) is to be removed from a cation, which has already acquired a noble gas configuration. On the other hand, in the alkaline Earth metal (Mg), the second electron is to be removed from a monovalent cation (Mg + ), which still has one electron..Solution:
Mass of the reactants = 10g Mass of the products = 4.6 + 5.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mas..
Solution
(a) Here the outermost shell corresponds to, n = 4. So, this element lies in the 4 t h period. (b) The outer electronic configuration of this element is 3d 1 0 4s 1 the energy of the '3d' orbitals is higher than that of '4s'. Therefore the last electron..
Solution:
22.4L of a gas = 1 GMM 100 cm 3 of gas = 0.48 g 22.4 x 1000 of gas = ? Gram molecular weight of the gas is 107.52 g. All elements are represented by symbols and all compounds represented by chemical formulae indicating the number of atoms of elements and also the p..
22.4L of a gas = 1 GMM 100 cm 3 of gas = 0.48 g 22.4 x 1000 of gas = ? Gram molecular weight of the gas is 107.52 g. All elements are represented by symbols and all compounds represented by chemical formulae indicating the number of atoms of elements and also the p..Solution
Mass number of 1 4 C is 14. So 14 g of 1 4 C contain 6.023 x 10 2 3 atoms of 1 4 C = 3.012 x 10 2 0 atoms Number of neutrons in 1 atom of 1 4 C = 14 - 6 = 8 Number of neutrons in 7 mg of 1 4 C..
Mass number of 1 4 C is 14. So 14 g of 1 4 C contain 6.023 x 10 2 3 atoms of 1 4 C = 3.012 x 10 2 0 atoms Number of neutrons in 1 atom of 1 4 C = 14 - 6 = 8 Number of neutrons in 7 mg of 1 4 C..Solution:
Initial velocity (u) = 0 Time taken (t) = 6 seconds Acceleration due to gravity (g) = 9.8 m/s 2 We make use of second equation of motion h = 0 + 4.9 x 36 h = 176.4 m 03. A stone projected vertically upward, takes 5 seconds to reach the highest..
Initial velocity (u) = 0 Time taken (t) = 6 seconds Acceleration due to gravity (g) = 9.8 m/s 2 We make use of second equation of motion h = 0 + 4.9 x 36 h = 176.4 m 03. A stone projected vertically upward, takes 5 seconds to reach the highest..Solution
Mass of the substance taken = 0.316 g Mass of BaSO 4 formed = 0.466 g From stoichiometry, BaSO 4 = S (molecular mass of the BaSO 4 = 137+ 32 + 64 = 233)233 32 Then, ..
Mass of the substance taken = 0.316 g Mass of BaSO 4 formed = 0.466 g From stoichiometry, BaSO 4 = S (molecular mass of the BaSO 4 = 137+ 32 + 64 = 233)233 32 Then, ..Solution:
put cosecx + cotx = t =- log t = - log |cosec x +cot x | \ The solution is y= log |cosec x+ cot x| + 2x+ c 2 Giving at x = p /2 = 0, y =..
put cosecx + cotx = t =- log t = - log |cosec x +cot x | \ The solution is y= log |cosec x+ cot x| + 2x+ c 2 Giving at x = p /2 = 0, y =..See what our Users say :
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