0.27 g of a long chain fatty acid was dissolved in `100cm^(3)` of hexane. `10ml` of this solution was added dropwise to the surface of water in a round watch glass . Hexane evaporates and a monolayeer is formed. The distance from edge to center of the watch glass is `10 cm` . What is the height of the monolayers? `["Density of fatty acid "=0.9 g cm^(-3),pi=3]`
0.27 g of a long chain fatty acid was dissolved in 100 cm^ 3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. The height of the monolayer is 10 ^( -x ) m. what is numerical value of x ? [Density of fatty acid = 0.9 g cm ^ ( -3) , pi = 3 ]
0.27 g of fatty acid is dissolved in 100 ml of solvent, 10 ml such solution is taken & placed over round plate. Distance from the centre of edge of round plate is 10 cm. Now solvent is evaporated & only fatty acid is remained. Density of fatty acid is 0.9g//c . Determine height of fatty acid layer. (pi=3)
A body weight 20 N in water and 30 N in air. Find the density of the body. Take the density of water as 1.0 g cm^(-3) .
A piece of gold weighs 10g in air and 9g in water. What is the volume of cavity? (Density of gold = 19.3 g cm^(-3) )
The depth of immersion of hydrometer in water is 10 cm. What will be the depth of immersion in a liquid whose density is 2 g cm^(-3) ?
An air bubble in a glass sphere (mu = 1.5) is situated at a distance 3 cm from a convex surface of diameter 10 cm . At what distance from the surface will be the bubble appear ?
An air bubble in a glass sphere (mu = 1.5) is situated at a distance 3 cm from a convex surface of diameter 10 cm . At what distance from the surface will the bubble appear ?
