Answer:
with water
water water water water water water water water 6
Answer:
It is present in third period that's why its valance electrons are present in 3rd energy level.
Its atomic number is greater than lithium when compared in group wise.
There are more electrons in sodium to shield the outer valance electron thus nuclear attraction becomes weak and size increase.
Explanation:
The size of sodium is greater than lithium because atomic number of sodium is 11 and lithium is 3. Both are present in first group but sodium is present down to the lithium. As we move from top to bottom in a group atomic size increases with addition of electrons. The nuclear effect become weaker on valance electrons and atomic size increase. Same time shielding effect is also produces which shield the outer electrons from the influence of nucleus. While in case of lithium less electrons are present to shield the valance electrons.
As we note the position of both elements along period. The sodium is present in third period while lithium is present in second period. So, in case of sodium third energy level is involved. That's why its size is greater than lithium.
Answer:
Wavelength of light in (nm) = 579 nm
Explanation:
At first you find out the amount of energy needed to just eject one electron. This is given by 
this energy is given in question in kj/mole. This is the work function of cesium for each electron is equal to the planc'k einstein equation.
Answer:
Mg.
Explanation:
- The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).
- An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced.
- A reducing agent (also called a reductant or reducer) is an element (such as calcium) or compound that loses (or "donates") an electron to another chemical species in a redox chemical reaction.
<em>Mg + 1/2O₂ → MgO.</em>
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Mg is oxidized to Mg²⁺ in (MgO) (loses 2 electrons). "reducing agent".
O is reduced to O²⁻ in (MgO) (gains 2 electrons). "oxidizing agent".
After careful consideration your answer is...
Leucippus and Democritus
*Hope I helped*
~Alanna~
