Benefits of soapy detergents over the soap less detergents
What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?
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Whenever we put on dirty clothes in the detergent then due to the chemical bonding made up of michell. One portion of michell termed head and the other part is termed as tail. Head comprises Na and tail comprises oH and whenever we dropped dirty cloth on water then it encompasses to be clean.
The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16) (a) 148 &nbs
Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre (d) Concentratio
Choose the right answer from following. If we take 44g of CO2 and 14g of N2 what will be mole fraction of CO2 in the mixture: (a) 1/5 (b) 1/3 (c) 2/3 (d) 1/4
Choose the right answer from following. Vapour pressure of a solution is: (a) Directly proportional to the mole fraction of the solvent (b) Inversely proportional to the mole fraction of the solute (c) Inversely proportional to the mole fraction of the solvent (d
arrange in order of basicity,pyridine,pipyridineand pyorine
Write down a short note on the differences between the organic and inorganic chemistry?
What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m (ii) 0.5m (iii) 0.2m (iv) 2m
Give me answer of this question. The volume of water to be added to 100cm3 of 0.5 N N H2SO4 to get decinormal concentration is : (a) 400 cm3 (b) 500cm3 (c) 450cm3 (d)100cm3
Select the right answer of the question. Molecular weight of urea is 60. A solution of urea containing 6g urea in one litre is : (a)1 molar (b)1.5 molar (c) 0.1 molar (d) 0.01 molar
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
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