<span>the formation of a gas
</span>
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Explanation:
When OH- (as in potassium hydroxide) is added, it reacts with the acid (HOCl) to reduce the amount of HOCl and increase the concentration of sodium hypochlorite.
Potassium hydroxide will react with the hypochlorous acid to produce hypochlorite ions. In the process, some of the weak acid will be consumed, along with the added strong base.
This occurs as follows:
HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)
since water is formed, this maintains the pH. Thus ...
A. The number of moles of HClO will decrease. - TRUE
B. The number of moles of ClO- will increase. - TRUE
C. The equilibrium concentration of H3O+ will remain the same. - TRUE
D. The pH will decrease. - FALSE
E. The ratio of [HClO] / [ClO-] will decrease. -TRUE. It will decrease as HClO goes down and ClO- goes up.
Answer:
The reaction when the Borane (BH3) is add to an alkene and form an alkylborane is shown below.
Explanation:
The boron of the borane does not have extra electron pairs, in this way the double bond of the alkene attacks the boron and the hydrogen belonging to the borane adheres to the carbon that is more substituted, thus forming an alkyl borane.
