Answer:
It's because removal of electron from an atom, reduces the size of an atom.
Explanation:
When an electron is removed from an atom, it becomes an ion and in this case it will become a postive ion.
When an electron is removed from an atom, the charge balance of an atom is disturbed and positive charge increases in comparison to the negative charge. This results in increase nuclear (positive) charge which exerts greater attraction on the remaining electrons and as a result the remaining electrons are more strongly attracted by the nucleus and in this way the atomic size is decreased. Due to this increased nuclear attraction and reduced atomic size, it bcomes difficult to remove more electeon from the positively charged ion of reduced size. This is the reason that each successive ionization of electron requires a greater amount of energy.
The ionization energy has inverse relation with the size or radius of an atom. This also justifies the reason that why each successive ionization of an electron requires greater amount of energy.
Answer:
the answer should be B xD<<<<<<<<<<<:::::::::::>>>>>>>>::::::::::::<<<<<<:::::>>>>>::::: makes this brainlist for good luck for the rest of ur life dont risk it
Explanation:
Molar mass H₂O = 18.0 g/mol
number of moles :
1.0 / 18.0 => 0.055 moles
1 mole -------------- 6.02 x 10²³ molecules
0.055 moles -------- ? molecules
molecules = 0.055 x ( 6.02 x 10²³) / 1
molecules = 3.311x10²² / 1
= 3.311 x 10²² molecules
hope this helps!
Answer:
0.0400 g for the example given below.
Explanation:
pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.
- By definition,
![pH = -log[H_3O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH_3O%5E%2B%5D)
. - NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation:
. - From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH:
![[NaOH] = [OH^-]](https://tex.z-dn.net/?f=%5BNaOH%5D%20%3D%20%5BOH%5E-%5D)
. - Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain:
![[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%20%3D%20%5BNaOH%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7Bm_%7BNaOH%7D%7D%7BM_%7BNaOH%7DV%7D)
. - We also know that
![pOH = 14.00 - pH = -log[NaOH]](https://tex.z-dn.net/?f=pOH%20%3D%2014.00%20-%20pH%20%3D%20-log%5BNaOH%5D)
. Take the antilog of both sides: ![10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}](https://tex.z-dn.net/?f=10%5E%7B-pOH%7D%20%3D%2010%5E%7BpH%20-%2014.00%7D%20%3D%20%5BNaOH%5D%20%3D%20%5Cfrac%7Bm_%7BNaOH%7D%7D%7BM_%7BNaOH%7DV%7D)
. - Solve for the mass of NaOH:
.
Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:
