<span>To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.</span>
Answer:
Explanation:
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In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:
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Explanation:
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Answer:
Group of highly-reactive chemical elements. The alkali metals are a group (column) in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr).
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.
