To prepare 350 mL of 0.100 M solution from a 1.50 M
solution, we simply have to use the formula:
M1 V1 = M2 V2
So from the formula, we will know how much volume of the
1.50 M we actually need.
1.50 M * V1 = 0.100 M * 350 mL
V1 = 23.33 mL
So we need 23.33 mL of the 1.50 M solution. We dilute it
with water to a volume of 350 mL. So water needed is:
350 mL – 23.33 mL = 326.67 mL water
Steps:
1. Take 23.33 mL of 1.50 M solution
<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M
solution</span>
Answer:
(a) Hypoeutectic
(b) Alpha solid, aluminium
(c) 70% α
, 30% β
(d) 97.6% α, 2.4% β
(e) 97.6% α, 2.4% β
(f) 97% α, 3% β
Explanation:
(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic
(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon
(c) At 578°C we have
% α: Al (11 - 4)/(11 - 1) = 70% α
% L: Si 100 - 70 = 30% β
(d) At 576°C we have
α: 99.83% Si (99.83 - 4)/(99.83- 1.65) = 97.6% α
β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β
(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α
Eutectic 4% Si = 100 - 97.6 = 2.4% β
(f) At 25°C we have;
α%: (99.83 - 4)/(99.83 - 1) = 97% α
β%: 100 - 97 = 3% β.
Answer:
It is the total distance traveled over the total time .
1. Pure substances cannot be separated into any other kinds of matter, while a mixture is a combination of two or more pure substances.
2. A pure substance has constant physical and chemical properties, while mixtures have varying physical and chemical properties (i.e., boiling point and melting point).
3. A pure substance is pure, while a mixture is impure.
Explanation:
The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)
The mass of aluminum present in the pot is:
Hence, in the given pot 270g Al is present.
The gram atomic mass of Al -27 g/mol
Given the mass of Al is 270 g
Substitute these values in the above formula:
Answer is 10.0 mol of Al is present.
