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.
Answer:
hy do C-C bonds and C-H bonds have high potential energy levels? ... electronegativity so the electrons in the covalent bond are equally shared. Term. Which functional groups are commonly found on a carbohydrate? ... few organisms have the specific enzymes needed to break the bond in the B-form ...
Explanation:
Gas x would be carbon dioxide.
note/ acid + carbonate —> salt + water + carbon dioxide
the white precipitate would be calcium carbonate. CaCo₃
note/ this is a common eqn u need to remember.
X - CO₂ (carbón dioxide)
Y - CaCo₃ (calcium carbonate)
sodium carbonate is a basic salt
The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
The energy of the photon is given by;
E = hc/λ
E = energy of the photon
h = Plank's constant
c = speed of light
λ = wavelength of light
For the upper boundary range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 270 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9
E = 7.33 × 10^-19 J
For the lower range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ =300 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9
E = 6.6 × 10^-19 J
Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
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