Explanation:
a) The amount of heat released by coffee will be absorbed by aluminium spoon.
Thus, 
To calculate the amount of heat released or absorbed, we use the equation:
Also,
![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=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
..........(1)
where,
q = heat absorbed or released
= mass of aluminium = 45 g
= mass of coffee = 180 g
= final temperature = ?
= temperature of aluminium = 
= temperature of coffee = 
= specific heat of aluminium = 
= specific heat of coffee= 
Putting all the values in equation 1, we get:
![45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]](https://tex.z-dn.net/?f=45%20g%5Ctimes%200.80J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-24%5EoC%29%3D-%5B180%20g%5Ctimes%204.186J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-83%5EoC%29%5D)
80.30 °C is the final temperature.
b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.
So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles = 
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = 
= 0.13moles
Number of moles of NH₃ = 
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
