C. Rubbing the balloon against your hair
Problem 2
You start out with 216 ugrams of Fermium - 253. After 3 days, you will have 1/2 as much. 108 ugrams is what you have.
Another 3 days goes by. You started with 108 ugrams. That gets cut in 1/2 again. Now you have 54 ugrams.
Finally another 3 days goes by. You started with 54 ugrams. you now have 1/2 as much which would be 27 ugrams
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.
<em><u>Table</u></em>
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
<em><u>Rules</u></em>
As the number of bonds INCREASES, the energy contained in the bond goes UP
As the number of bonds INCREASES, the length of the bond goes DOWN.
Synapse / Neuronal Junction
This can be done in the following way;
1 determining the heat required to convert 0° C ice to 0°C water
Heat of fusion of water = 334 J/g
Therefore; Heat = 50 g × 334 J/g = 16700 J
2. Determining the heat required to raise the temperature of water from 0° C to 100°C.
Specific heat of water is 4.18 J/g°C
Change in temperature is 100°C
Therefore; Heat = 50 g × 4.18 J/g°C × 100 = 20900 J
3. Determining the heat required to convert 100 ° C water to 100°C vapor
Heat of vaporization of water = 2257 J/g
Heat = Mass of water × heat of vaporization
Heat = 50 g × 2257 = 112850 J
4. Determining the heat required to go from 100° C to 120° vapor
specific heat of vapor = 2.09 J/g°C
Heat = mass × Specific heat of vapor × change in temperature
= 50 g × 2.09 ×(120-100) = 2090 J
Therefore the total heat required is
= 16700 J + 20900 J + 112850 J + 2090 J = 152540 J or 152.54 kJ
