Ionic compounds are those compounds which are made up of ions. These ions can exist as positive or negative ions. Those ions which has tendency to lose electrons are said to be cations (positive ions) such as metals and those ions which has tendency to gain electrons are said to be anions (negative ions) such as non-metals.
The formula for ionic compounds are given as:
First, the symbols of the cation and anion should be identified. After that, charge on the cation and anion should be identified. Write the chemical formula with the help of drop and swap method.
Now,
The chemical symbols for Aluminium is 
and for chlorine is 
.
The charge on is +3 and on is -1, neglect the positive and negative signs and by using drop and swap method i.e. the charge on the aluminium will go as a subscript of the chlorine and the charge on the chlorine will go as a subscript of the aluminium which is shown in the image.
Thus, the chemical formula between and is 
.
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Answer:1)1.99 M
Explanation:
Molarity is given as = moles solute/Liter solution
The solute which is LiOH is already given in moles as 1.495 moles
Given that solution is 750 mL, we convert to liters.
Liters of the solution= mL of the solution x (1 L/1000 mL)
750 mL x (1 L/1000 mL)
0.75 L
Molarity = moles solute/Liter solution
Molarity = 1.495 moles of LiOH/0.75 L of solution
Molarity = 1.99M
The molarity of this solution is 1.99M (moles per liter).
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Answer:
a) Acidic buffer
b) No buffer
c) Acidic buffer
d) Basic buffer
e) Basic buffer
Explanation:
a) 75.0 mL of 0.10 M HF ; 55.0 mL of 0.15 M NaF
-Acidic buffer
Mixing of 75.0 mL of 0.10 HF and 55.0 mL of 0.15 mL NaF results in acidic buffer. HF/NaF is a buffer of weak acid and its conjugate base. F- is the conjugate base of acid,HF.
b.) 150.0 mL of 0.10 M HF ; 135.0 mL of 0.175 M HCl-No buffer
Mixing HF and HCl will not results in a buffer. Both are acids, and no conjugate base is present.
c.) 165.0 mL of 0.10 M HF ; 135.0 mL of 0.050 M KOH-Acidic buffer
HF reacts with KOH to form KF. F- is a conjujate base of HF. As volume and concentration of HF is more than KOH, therefore, HF will remain after reaction with KOH. HF/KF will be a buffer of weak acid and its conjugate base.
d.) 125.0 mL of 0.15 M CH3NH2 ; 120.0 mL of 0.25 M CH3NH3Cl -Basic buffer
CH3NH2/CH3NH3+ is a buffer of weak base and its conjugate acid.
e.) 105.0 mL of 0.15 M CH3NH2 ; 95.0 mL of 0.10 M HCl-Basic buffer
CH3NH2 is a weak base and HCl is a strong acid. CH3NH2 reacts with HCl to form its conjugate acid CH3NH3+. Volume and concentration of CH3NH2 is more as compared to HCl and hence, will remain in the soution after reactionf with HCl.
CH3NH3+/CH3NH2 is a buffer of weak base and its conjugate acid.
