Conditions for rusting
• moisture
•oxygen
Prevention
•Painting
• Galvanisation
• Making Alloys
• greasing
Hope this helps you ;)(:
Boiling Point, Melting Point, Viscosity, Surface Tension. Decrease: Vapor Pressure.
You could use a scale to measure the mass as well as a cup to hold the water. If you were comparing the two, you should also probably use a graduated cylinder to get the same amount of each type of water.
Hope this helped ^_^
Answer:
132g/mole
Explanation:
using the formula PV=nRT should be used to solve for the number of moles (n). R is a constant which is 62.3637 L mmHG/mole K.
Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K. Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.
