Calculation
Mass of U-tube with contents before experiment = a grams Mass of U-tube with contents after experiment = b grams Hence, mass of water vapour absorbed = (b - a) g Volume of air = x litres..
Mass of U-tube with contents before experiment = a grams Mass of U-tube with contents after experiment = b grams Hence, mass of water vapour absorbed = (b - a) g Volume of air = x litres..Calculations
Let, The mass of the compound taken be = W g Mass of AgX formed = W 1 g From stoichiometry, Ag = X 1 mol 1 mol 1 molecular mass 1 atomic mass Therefore, ..
Let, The mass of the compound taken be = W g Mass of AgX formed = W 1 g From stoichiometry, Ag = X 1 mol 1 mol 1 molecular mass 1 atomic mass Therefore, ..Calculations
Let, The mass of the organic compound taken be = W g Mass of Mg 2 P 2 O 7 obtained = W 1 g From stoichiometry, Mg 2 P 2 O 7 = 2P 1 mol 2 mol 1 x molecular mass 2 x atomic mass 222 g 2 x 31 g = 62 g Therefore, Then, ..
Let, The mass of the organic compound taken be = W g Mass of Mg 2 P 2 O 7 obtained = W 1 g From stoichiometry, Mg 2 P 2 O 7 = 2P 1 mol 2 mol 1 x molecular mass 2 x atomic mass 222 g 2 x 31 g = 62 g Therefore, Then, ..Calculations
Let the mass of platinic chloride salt be = W g The mass of platinum residue left = x g From the molar stoichiometry, Number of mol of B 2 H 2 PtCl 6 = Number of mol of Pt Or If E is the equivalent mass of the organic base, Then molecular mass of the salt (B 2 H 2 PtCl 6 ) = 2E + 2 + 195 + 213 ..
Let the mass of platinic chloride salt be = W g The mass of platinum residue left = x g From the molar stoichiometry, Number of mol of B 2 H 2 PtCl 6 = Number of mol of Pt Or If E is the equivalent mass of the organic base, Then molecular mass of the salt (B 2 H 2 PtCl 6 ) = 2E + 2 + 195 + 213 ..Calculations
Let, The mass of the acid dissolved be = W g Volume of alkali used for complete neutralization of acid solution = V 1 mL Normality of the alkali solution = N 1 Then, we can write, V 1 mL of alkali solution having normality N 1 = Wg acid From the definition, 1000 mL of 1N alkali solution = 1..
Let, The mass of the acid dissolved be = W g Volume of alkali used for complete neutralization of acid solution = V 1 mL Normality of the alkali solution = N 1 Then, we can write, V 1 mL of alkali solution having normality N 1 = Wg acid From the definition, 1000 mL of 1N alkali solution = 1..Calculation
Let, Mass of the organic compound = W g Volume of the standard acid required for complete neutralization of the evolved ammonia = V mL Normality of the standard solution of acid = N From the law of equivalence (normality equation), 1000 mL of 1 N acid = 1000 mL of 1 N NH 3 = 17g NH 3 = 14 g ni..
Let, Mass of the organic compound = W g Volume of the standard acid required for complete neutralization of the evolved ammonia = V mL Normality of the standard solution of acid = N From the law of equivalence (normality equation), 1000 mL of 1 N acid = 1000 mL of 1 N NH 3 = 17g NH 3 = 14 g ni..Calculation of valency
The number of valence electrons is the valency of the element. The valency of an element can also be calculated by finding the number of electrons required to complete octet (8). i.e., 8 number of valence electrons = valen..
Calculation of Valency
The number of valence electrons is the valency of the element. The valency of an element can also be calculated by finding the number of electrons required to complete octet (8)i.e., 8 - Number of valence electrons = Valenc..
We ____ the exponents to divide exponential expressions with same bas..
We ____ the exponents to divide exponential expressions with same base. => multiply or add or subtract or divide..
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