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.
The volume of 0.20 moles of helium at STP is 4.5 liters.
Explanation:
Given:
Number of moles = 0.20 moles
To Find:
The volume of Helium at STP =?
Solution:
According to ideal gas law
PV = nRT
where
P is pressure,
V is volume,
n is the number of moles
R is the gas constant, and
T is temperature in Kelvin.
The question already gives us the values for p and T
,because helium is at STP. This means that temperature is 273.15 K and pressure is 1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol since these units fit the units of our given values the best
On substituting these values we get



V = 4.5 Liters
At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D
I believe doctors use (3) Co-60 to treat ILWs. Uranium and Carbon are not used, and a different isotope of lead is used to be attached to monoclonal antibodies.