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,
..........(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:
hope it helps you
Explanation:
Once one shell is full, the next electron that is added has to move to the next shell. So... for the element of NEON, you already know that the atomic number tells you the number of electrons. That means there are 10 electrons in a neon atom.
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M