The % mass/mass concentration of solute in the seawater sample is calculated as below
% mass = mass of the solute /mass of the solvent(sea water) x100
mass of the solute =1.295 g
mass of the solvent(sea water_) = 25.895 g
there the % mass = 1.295/25.895 x100 = 5.001 %
Answer: The differences is Air masses cover over thousands and hundreds and millions of square kilo. A front is a boundary which two air masses but different temperature and moisture content meet.
Explanation:
Answer:
0.0014 moles is present in 40cm³ of 0.035M of HCl solution
Explanation:
Molarity = 0.035M
V = 40.0mL
1mL = 1cm³
V = 40cm³
0.035 moles = 1000cm³
X moles is present in 40cm³
X = (40 * 0.035) / 1000
X = 0.0014moles
0.0014 moles is present in 40cm³ of solution
Because they do not have the same qualities therefore they are different
