Answer: The mass of aluminium sample is 55.4 gram
Explanation:
As we know that,
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
where,
= mass of aluminium = ?
= mass of water = 300.0 g
= final temperature = 
= temperature of aluminium = 
= temperature of water = 
= specific heat of aluminium = 
= specific heat of water = 
Now put all the given values in equation (1), we get
![-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=-%5Bm_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%5D%3D%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
![-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)]](https://tex.z-dn.net/?f=-%5Bm_1%5Ctimes%200.897%5Ctimes%20%2823.8-94.5%29%5E0C%5D%3D%5B300.0g%5Ctimes%204.184%5Ctimes%20%2823.8-21.0%29%5D)
Therefore, the mass of aluminium sample is 55.4 gram
2 resonance structure.
O=S⁺-O⁻ and O⁻-S⁺=O, angles between S and O are 120°, bond order is 1,5.
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.
Idk how to write the formula I had this same question
A) Fe⁰ ----> Fe⁺³ +3e⁻ oxidation | *2
b) <u>Cu⁺² + 2e⁻ -----> Cu⁰ reduction |*3</u>
c) 2Fe⁰ +3Cu⁺² -----> 2Fe⁺³ + 2Cu⁰
