Answer:
B. A chemist
<em>Hope it's help :)</em>
Answer:
800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Explanation:
3000 lb of 13% solution is required .
Total adhesive in weight = 3000 x .13 = 390 lb of adhesive
Available = 500 lb of 10% solution = 50 lb of adhesive
Rest = 390 - 50 = 340 lb required .
rest mass of solution = 3000 - 500 = 2500 lb
mass of adhesive required = 340 lb
Let the mass of 20% required be V
mass of adhesive = .20 V
.20 V = 340
V = 1700
rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent
So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
<span><span>Yes.
An element that is highly electronegative pulls more on the electrons
in a bond, such as oxygen in H20. This creates a polar bond, where
there is a small negative charge on the oxygen, and a small positive
charge in between the hydrogens.
</span>Credit goes to "Erin M" answered on yahoo answers a decade ago.
</span>
Answer:
If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.
Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.
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.
