Number of electron pairs = \frac{1}{2}[V+N-C+A]
2
1
[V+N−C+A]
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
SbCl_5SbCl
5
:
In the given molecule, antimony is the central atom and there are five chlorine as monovalent atoms.
The number of electron pairs are 5 that means the hybridization will be sp^3dsp
3
B and geometry of the molecule will be trigonal bipyramidal.
Following are important constant that used in present calculations
Heat of fusion of H2O = 334 J/g
<span>Heat of vaporization of H2O = 2257 J/g </span>
<span>Heat capacity of H2O = 4.18 J/gK
</span>
Now, energy required for melting of ICE = <span> 334 X 5.25 = 1753.5 J .......(1)
Energy required for raising </span><span>the temperature water from 0 oC to 100 oC = 4.18 X 5.25 X 100 = 2195.18 J .............. (2)
</span>Lastly, energy required for boiling water = <span> 2257X 5.25 = 11849.25 J ......(3)
</span><span>
Thus, total heat energy required for entire process = (1) + (2) + (3)
= 1753.5 + 2195.18 + 11849.25
= </span><span>15797.93 J
</span><span> = 15.8 kJ
</span><span>Thus, 15797.93 J of energy is needed to boil 5.25 grams of ice.</span>
Answer: The new volume at different given temperatures are as follows.
(a) 109.81 mL
(b) 768.65 mL
(c) 18052.38 mL
Explanation:
Given:
= 571 mL, ![T_{1} = 26^{o}C](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%2026%5E%7Bo%7DC)
(a) ![T_{2} = 5^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%205%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B5%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%20109.81%20mL)
(b) ![T_{2} = 95^{o}F](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%2095%5E%7Bo%7DF)
Convert degree Fahrenheit into degree Cesius as follows.
![(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C](https://tex.z-dn.net/?f=%281%5E%7Bo%7DF%20-%2032%29%20%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%20%3D%20%5E%7Bo%7DC%5C%5C%2895%5E%7Bo%7DF%20-%2032%29%20%5Ctimes%20%5Cfrac%7B5%7D%7B9%7D%20%3D%2035%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B35%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%20768.65%20mL)
(c) ![T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%201095%20K%20%3D%20%281095%20-%20273%29%5E%7Bo%7DC%20%3D%20822%5E%7Bo%7DC)
The new volume is calculated as follows.
![\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B1%7D%7D%7BT_%7B1%7D%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7BT_%7B2%7D%7D%5C%5C%5Cfrac%7B571%20mL%7D%7B26%5E%7Bo%7DC%7D%20%3D%20%5Cfrac%7BV_%7B2%7D%7D%7B822%5E%7Bo%7DC%7D%5C%5CV_%7B2%7D%20%3D%2018052.38%20mL)
A logarithmic scale is a nonlinear scale used when there is a large range of quantities