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Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M
30 / 1000 L = 0.140 M
0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2
( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol
( 1 L / 0.570 mol )
( 1000 mL / 1 L ) = 7.37 mL of KOH
The answer to the question Retraction. Its because the light goes around it
The answer would be “adding the sum of”
Answer:
pOH = 5
pH = 9 → Solution is basic
Explanation:
We need to know that:
pH + pOH = 14
< 7 ⇔ pH = 7 ⇔ > 7
Acidic Neutral Basic
pOH = - log [OH⁻] → - log 9.8×10⁻⁶ = 5.00
pOH = 5
pH = 14 - 5 = 9