Argon atoms would not tend to make bonds with other atoms since it is classified as a noble atom which means it has a complete filled subshells. It does not have any electrons that are available for bonding. Noble gases belong to the Group 8A in the periodic table of elements. All of the elements in this group do have a complete or filled electron orbitals which makes them so stable or the s and p outer electron shells are complete with electrons. All of them follows the octet rule. Thus, they do not need any sharing or transfer of electrons from other compounds for them to be more stable.
Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
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Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
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Given:
Concentration 1 = 100.0 ml Concentration 2 = ?
Volume 1 = 0.100 M KCL Volume 2= 0.500 M KCL
Find Concentration 2
To get the concentration 2, follow this formula
C1V1 = C2V2
Solution
(100.0 ml) (0.100 m kcl) = (X ml) (0.500 m kcl)
transfer the volume 2 to the left side
(100.0 ml) (0.100 m kcl) = X ml
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0.500 m kcl
10.00ml / m kcl = Xml
-------------------- cancel the m kcl to let the ml be the remaining variable
0.500 m kcl
20.0 ml = x ml
So you will need 20.0 ml of 0.500 m kcl to made the solution of 100.0 ml of 0.100 m kcl
