The molarity of formic acid is 100 mM or 
. The dissociation reaction of formic acid is as follows:
The expression for dissociation constant of the reaction will be:
![K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BHCOOH%5D%7D)
Rearranging,
![[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7BK_%7Ba%7D%5BHCOOH%5D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:
![[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-4.15%7D%3D7.08%5Ctimes%2010%5E%7B-5%7DM)
Similarly, 
thus,
Putting the values,
![[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M](https://tex.z-dn.net/?f=%5BHCOO%5E%7B-%7D%5D%3D%5Cfrac%7B%281.78%5Ctimes%2010%5E%7B-4%7DM%29%28100%5Ctimes%2010%5E%7B-3%7DM%29%7D%7B%287.08%5Ctimes%2010%5E%7B-5%7DM%7D%3D0.2511%20M)
Therefore, the concentration of formate will be 0.2511 M.
False cause there particles arent large enough to be filtered
Well, an element is a substance that cannot be decomposed into a simpler substance by chemical change, so element is the answer you are probably looking for! Hope this helps!
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The compound's molecular formula is C2H6. This is obtained by:
mass moles divided by smallest moles
C 32g 32/12 = 2.67 1
H 8g 8/1.01 = 7.92 approx. 3
Next, divide both terms by the smallest number of moles, 2.67. This gives 1 and 3. So the empirical formula is CH3 which has a molar mass of 15g/mol. Given the molar mass of the molecular formula as 30g/mole, we can calculate the factor by which to multiply the subscripts of CH3.
X = molar mass of molecular formula / molar mass of empirical formula = 30/15
X=2
So (CH3)2 is C2H6.
