Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
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Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)
Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is
0,00014 M
Have a nice day!
The mass of the piece of lead is calculated using the below formula
Q(heat)= mC delta T
Q = 78.0 j
M=mass =?
C=specific heat capacity ( 0.130 j/g/c
delat T=change in temperature = 9.0 c
by making M the subject of the formula
M = Q/ c delta T
M= 78.0 j/ 0.130 j/g/c x 9.0 c = 66.7 g of lead
Answer:
I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.
Explanation:
<em>I verify this is correct. </em>
When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺
