Answer:
The answer is "Option b and Option c".
Explanation:
This buffer is a buffer of ammonia and ammonium ion. Thus it requires the solution 
.
In point 1:
The solution containing at 1M concentration would be given by mixing the two solutions. Thus, this buffer is a legitimate route.
In point 2:
It gives the ions you want but they are not the same.
In point 3:
and 
volume would not produce the same concentrations. Therefore, this buffer isn't a valid route.
In point 4:
Some volume and half . This offers the same rate as half.
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The Options are as follow,
<span> (1) CaCl</span>₂<span> (s) (3) CH</span>₃<span>OH (l)</span>
<span> (2) C</span>₂<span>H</span>₆<span> (g) (4) Cal</span>₂<span> (aq)</span>
Answer:
Option-1 is the correct answer.
Explanation:
As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.
Option A, CaCl₂ is in a solid state , so it can exist in crystal form.
Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.
Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.
Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.
The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.
Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).
Nomenclature:
In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.
Name of Formate:
Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.
E.g.
HCOOH (formic acid) → HCOO⁻ (formate) + H⁺
H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺
Formal Charges:
Formal charges are calculated using following formula,
F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]
For Oxygen:
F.C = [6] - [6 + 2/2]
F.C = [6] - [6 + 1]
F.C = 6 - 7
F.C = -1
For Sodium:
F.C = [1] - [0 + 0/2]
F.C = [1] - [0]
F.C = 1 - 0
F.C = +1
