It is both. It is a mixture because it has two or more things mixed in it. It is a solution, because all of the components and things mixed in the milk are uniform and milk isn't chunky or discolored because of that.
Hope this helps.
Answer #1. A 2.5% (by mass) solution concentration signifies that there is 2.5 grams of solute in every 100 g of solution.
To calculate 2.5% by mass solution, we divide the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2. therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution.
This means that weighing out 2.5 grams of solute and then adding 97.5 grams of solvent would make a total of 100 gram solution:
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3. a solution mass of 1 kg is 10 times greater than 100 g, thus 1kg of a 2.5% ki solution would contain 25 grams of ki.
Since 1000 grams is 1 kg, we multiply 10 to each mass so that 100 grams becomes
1000grams:
mass of solute / mass of solution = 2.5g*10 / [(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
The Antimony Hexachloride Formula includes:
Answer:
(a). 132 × 10^-9 s = 132 nanoseconds.
(b)..176.5 pico-seconds.
Explanation:
(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;
Speed = [ (8 × R × T)/ Mm × π]^1/2.
Where Mm = molar mass, T = temperature and R = gas constant.
Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.
The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;
Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.
Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.
Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.
Degree of collision = 7.55 × 10^6 s^-1.
Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.
(b). At one bar;
1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.
