Explanation:
Physical change -
It refers to any change during any process , where there is no new substance is formed , is referred to as a physical change.
Any change in the state of matter is characterised under physical change.
Chemical change -
It refers to any change in a chemical process , where there is formation of any new substance , is referred to as a chemical change .
From the options given in the question,
1. Apple turning brown , on air exposure , is an example of chemical change .
2. The physical state of mercury is liquid , is an example of physical change .
3. phosphorus burns on exposed to air , is an example of chemical change .
4. The gas Neon is colorless at the room temperature , is is an example of physical change .
Answer:
35.4528731 amu
Explanation:
To appropriately get the atomic mass unit of chlorine, we can get the answer using the masses from the isotopes. This can be obtained as follows. What we do is that we multiply the percentage compositions by the masses.
Now let’s do this.
[75.77/100 * 34.969] + [24.23/100 * 36.966]
= 26.4960113 + 8.9568618 = 35.4528731
To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:
Heat = -3.16 (64/32.06) = - 6.32 kJ
Hope this answers the question.
Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:
Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m
Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)
Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= 
The enthalpy change, ΔH, for this reaction per mole of X:
Answer : The amount of heat released, 45.89 KJ
Solution :
Process involved in the calculation of heat released :
Now we have to calculate the amount of heat released.
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+\Delta H_{fusion}+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
Q = amount of heat released = ?
m = mass of water = 27 g
= specific heat of liquid water = 4.184 J/gk
= specific heat of solid water = 2.093 J/gk
= enthalpy change for fusion = 40.7 KJ/mole = 40700 J/mole
conversion : 
Now put all the given values in the above expression, we get
![Q=[27g\times 4.184J/gK\times (314-273)k]+40700J+[27g\times 2.093J/gK\times (273-263)k]](https://tex.z-dn.net/?f=Q%3D%5B27g%5Ctimes%204.184J%2FgK%5Ctimes%20%28314-273%29k%5D%2B40700J%2B%5B27g%5Ctimes%202.093J%2FgK%5Ctimes%20%28273-263%29k%5D)
(1 KJ = 1000 J)
Therefore, the amount of heat released, 45.89 KJ
