<u>Answer:</u> The 
of the acid is 6.09
<u>Explanation:</u>
For the given chemical reaction:
The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:
![K_a=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
We are given:
![[HA]_{eq}=0.200M](https://tex.z-dn.net/?f=%5BHA%5D_%7Beq%7D%3D0.200M)
![[H^+]_{eq}=4.00\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D_%7Beq%7D%3D4.00%5Ctimes%2010%5E%7B-4%7DM)
![[A^-]_{eq}=4.00\times 10^{-4}M](https://tex.z-dn.net/?f=%5BA%5E-%5D_%7Beq%7D%3D4.00%5Ctimes%2010%5E%7B-4%7DM)
Putting values in above expression, we get:
p-function is defined as the negative logarithm of any concentration.
So,
Hence, the of the acid is 6.09
The answer is 'lipids'
hopefully this helps
Answer:
M = 0.441 M
Explanation:
In this case, we have two solutions that involves the Manganese II cation;
We have Mn(CH₃COOH)₂ and MnSO₄
In both cases, the moles of Mn are the same in reaction as we can see here:
Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻
MnSO₄ <------> Mn²⁺ + SO₄²⁻
Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:
moles of MnAce = 0.489 * 0.0283 = 0.0138 moles
moles MnSulf = 0.339 * 0.0125 = 0.0042 moles
the total moles are:
moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles
Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L
M = 0.018 / 0.0408
M = 0.441 M
This would be the final concentration of the manganese after the mixing of the two solutions
Bones are made up of a framework of a protein called collagen , with a mineral called calcium phosphate that makes the framework hard and strong. Bones store calcium and release some into the bloodstream when it's needed by other parts of the body.
