Answer:
The molarity of the base is 0.2664 M
Explanation:
Firstly, we write the complete and balanced titration reaction.
+ 
→ 
+ 
From the reaction, we can identify that 1 mole of the acid reacted with 2 moles of the alkali (base) to yield salt and water only
We identify the following also from the question;
= 14.80 mL , 
= 25.00 mL , 
= 0.225M and 
= ?
= 1 , 
= 2
We use the relation;
/ = /
Plugging the values, we have ;
(0.225 × 14.80)/1 = ( × 25.00)/2
= (2 × 0.225 × 14.80)/25
= 0.2664 M
Well you see its part sodium and sodium is another word for salt so It is a salt because it is formed by the reaction of an acid and a base
“The trapping of the suns warmth in a planets lower atmosphere, due to the greater transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planets surface.”
Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :
where,
A = absorbance of solution
C = concentration of solution
l = path length
= molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:
Given:
= 0.350
= 0.460
= 0.135 M
= ?
Now put all the given values in the above formula, we get:
Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
