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:
Option B is correct: C6H1206(s) and HCl(g)
Calculate the H positive from the pH equation: pH equals -log (H positive). This would be 10 to the -6.49. Let's call the acid HA. To calculate Ka in this equation, Ka equals H positive times A- over HA. HA is going to be the 0 0121. So, Ka=(10^-6.49)^2/0.0121. This equals 1.05*10^-13/0.0121. Ka then equals 8.65*10^-12.
Answer:
![AU^{3+} : [Rn] 5f^3](https://tex.z-dn.net/?f=AU%5E%7B3%2B%7D%20%3A%20%5BRn%5D%205f%5E3)
Explanation:
Writing electronic configuration of any element you should know atomic number of that element ,
and also electrons are filling according to their energy level and first electron is filled in the lower energy orbital
and it follows n+1 rule if n+1 is same for two orbital electron will go first in the lowest value of n.
writing electronic configuration of ion can be done like first for their neutral atom and then add or remove electron it will make things easy because there are also some eception case their you may do wrong.
![AU : [Rn] 5f^3 6d^1 7s^2](https://tex.z-dn.net/?f=AU%20%3A%20%5BRn%5D%205f%5E3%206d%5E1%207s%5E2)
remove three electron from outer most shell of AU
