What do budding, binary fishing and spore formation have in commen?
1 answer:
You might be interested in
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
Answer:
1.40 moles.
Explanation:
The balanced chemical equation for the reaction of Phosphorus and Oxygen is as follows -
In Part A, the oxygen was taken in excess. So Phosphorus will be the limiting reagent.
Since, 2 moles of is formed by 4 moles of
So, for 1.8 moles of amount of required moles of
=
In Part B, the phosphorus was taken in excess so oxygen will be the limiting reagent.
Since, 2 moles of is formed by 5 moles of oxygen
So, for 1.40 moles of moles of
required =
Thus as of now we have 3.60 moles of and 3.50 moles of
.
As in the reaction of formation of , oxygen is the limiting reagent.
So the moles of formed by the 3.50 moles of oxygen will be
=
= 1.40 moles.