Answer:
25.907°C
Explanation:
In Exercise 102, heat capacity of bomb calorimeter is 6.660 kJ/°C
The heat of combustion of benzoic acid is equivalent to the total heat energy released to the bomb calorimeter and water in the calorimeter.
Thus:

= heat of combustion of benzoic acid
= heat energy released to water
= heat energy released to the calorimeter
Therefore,
![-m_{combust}*H_{combust} = [m_{water}*c_{water} + C_{calori}]*(T_{f} - T_{i})](https://tex.z-dn.net/?f=-m_%7Bcombust%7D%2AH_%7Bcombust%7D%20%3D%20%5Bm_%7Bwater%7D%2Ac_%7Bwater%7D%20%2B%20C_%7Bcalori%7D%5D%2A%28T_%7Bf%7D%20-%20T_%7Bi%7D%29)
1.056*26.42 = [0.987*4.18 + 6.66](
- 23.32)
27.8995 = [4.12566+6.660](
- 23.32)
(
- 23.32) = 27.8995/10.7857 = 2.587
= 23.32 + 2.587 = 25.907°C
<span>c = speed of light = 3.00 x 10^5 km/s = 3.00 x 10^8 m/s
λ = wavelength of the microwave radiation = 3.50 cm = 0.035 m
f = frequency (in Hertz) = to be determined
f = c/λ = 3.00 x 10^8 m/s / 0.035 m
f = 8.57 x 10^9 Hz Frequency</span>
The answer would be B. One region of the molecule has a small negative charge while another region has a small positive charge. However usually in polar bonds, charges or bond between the atoms are unequal (as opposed to having small equal charges).
Explanation:
Upon dissolution of KCl heat is generated and temperature of the solution raises.
Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

= 4.31 kJ
or, = 4310 J (as 1 kJ = 1000 J)
Mass of solution will be the sum of mass of water and mass of KCl.
Mass of Solution = mass of water + (no. of moles of KCl × molar mass)
= 200 g + 
= 200 g + 13.625 g
= 213.625 g
Relation between heat, mass and change in temperature is as follows.
Q = 
where, C = specific heat of water = 
Therefore, putting the given values into the above formula as follows.
Q = 
4310 J =
Thus, we can conclude that rise in temperature will be
.
The majority of the mass of the atom is located in the nucleus. Remember that the nucleus contains both protons and neutrons and therefore, most of the mass of the atom.