The answer is B: multiple orbitals of specific magnitude within an energy level
Answer:
mass of water, mw = 300g = 0.3kg
∆Tw = (80 - 20) °C
volume of air, va = 100m³
mass of air, ma = 100g = 0.1kg
∆Ta = ?
H = mc∆T
Hw = mwcw∆Tw
Hw = 0.3*4200*60
Hw = 75600J
Hw = 75.6 kJ
All the above heat energy got absorbed by air,
that is; Ha = 75600J
since it's given that the heat was absorbed by a specific amount of volume of air
then specific capacity of volume of air is
then,
ca = <u>Ha</u><u> </u> × <u>density</u><u> </u>
ma temp
then,
Ha = vaca∆Ta
where, ca = volumetric heat capacity of air = 0.012kJ/m³°C
75.6kJ = 100m³ × 0.012kJ/m³°C × ∆Ta
75.6 = 1.2/°C × ∆Ta
∆Ta = 63°C
63°C is the temperature change in air.
Answer : The specific heat of the metal is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of metal = ?
= specific heat of water = 
= mass of metal = 50.0 g
= mass of water = 125 g
= final temperature of mixture = 
= initial temperature of metal = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
![(50.0g)\times c_1\times (29.3-115.0)^oC=-[(125g)\times 4.18J/g^oC\times (29.3-25.6)^oC]](https://tex.z-dn.net/?f=%2850.0g%29%5Ctimes%20c_1%5Ctimes%20%2829.3-115.0%29%5EoC%3D-%5B%28125g%29%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%2829.3-25.6%29%5EoC%5D)
Therefore, the specific heat of the metal is,
You need to find the formula mass of NaCl first, which you find the amu of both sodium and chlorine on the periodic table and you add them both. Then you divide the 175.5g by the formula mass which I got 58 g to get 3.026 moles of NaCl. 4 sig figs because of the 175.5. I recommend doing it on your own to make sure you get the right answer. You divide the 175.5g of NaCl by the formula mass.
Descriptive, Comparative, and experimental.
