Answer:
that's because....
group 1 (e.g Na, K) those tend to lose one electron to gain noble gas electron configuration.
they can achieve that by just losing one electron from their outer shell.
as you go down the group 1, element gets bigger in size, which means there is more space between nucleus (which is in center of atom) and electron of outer shell. the more far away they are the less attraction force between them.
so its easier for potassuim to lose one electron than for lithuim.
so that means potassium will easily give up 1 electron to react with non metal or other element therefore it is more reactive than lithuim
but in case of non metal, the opposite happens but simple to understand.
as you go down the group 7 (halogen- Cl, Br, I) element will get bigger therefore force between nucleus and outer electron is getting smaller. they have to gain 1 electron in order to fill the outer shell (to gain noble gas electron configuration.)
as florine is more smaller in size than clorine it is more reactive because florine has more tendency to pull extra electron from metal or other element towards its side. so it easily gain 1 electron to react.
This is what a chromosome looks like during mataphase
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
You are looking at a decomposition reaction because the original compound is being broken into its individual elements.<span />
