H. The atom would no longer be aluminum
If you added a proton to an atom of aluminum it would become a silicone ion
Answer:
second energy level
Explanation:
Valence electrons are those electrons which are present in outer most orbital of the atom.
This can be easily found through the electronic configuration of atom.
Electronic configuration of F:
F₉ = 1s² 2s² 2p⁵
We can see that the valence electrons are present in second energy level of F atom.
There are seven valence electrons of fluorine.
It is called halogens.
Halogens are very reactive these elements can not be found free in nature. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while iodine is solid.
Fluorine:
1. it is yellow in color.
2. it is flammable gas.
3. it is highly corrosive.
4. fluorine has pungent smell.
5. its reactions with all other elements are very vigorous except neon, oxygen, krypton and helium.
Answer:
V/V% = 8.2%
Explanation:
Given data:
Volume of methanol = 37.5 mL
Volume of solution = 456 mL
V/V% = ?
Solution:
V/V% = [volume of solute / volume of solution ]×100
V/V% = 37.5 mL / 456 mL × 100
V/V% = 0.08× 100
V/V% = 8.2%
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
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<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
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<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
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<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
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<em>C. at low temperature and low pressure.</em>
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