N(CuSO₄)=135/{63.5+32.1+4*16.0}=0.846 mol
n(CuSO₄*5H₂O)=135/{63.5+32.1+4*16.0+5*(2*1.0*+16.0)}=0.541 mol
n(H₃PO₄)=135/{3*1.0+31.0+4*16.0}=1.378 mol
n(BaCl₂)=135/{137.3+2*35.5}=0.648 mol
<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is 
We are given:
Putting values in above equation, we get:
Hence, the molarity of in the solution is 0.1 M.
Hey there, The answer is A. Free electrons.
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
