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:
760 mm of Hg
Explanation:
If the gases A , B and C are non reacting , then according to <u>Dalton's </u><u>Law </u><u>of</u><u> </u><u>Partial </u><u>Pressure</u> the total pressure exerted is equal to sum of individual partial pressure of the gases .
If there are n , number of gases then ,
Here ,
- Partial pressure of Gas A = 400mm of Hg
- Partial pressure of Gas B = 220 mm of Hg
- Partial pressure of Gas C = 140mm of Hg
Hence the total pressure exerted is ,
Substitute ,

Add ,

<u>Hence</u><u> the</u><u> </u><u>total</u><u> pressure</u><u> exerted</u><u> by</u><u> the</u><u> </u><u>gases </u><u>is </u><u>7</u><u>6</u><u>0</u><u> </u><u>mm </u><u>of </u><u>Hg</u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em>.</em>
Waters boiling point is 100°C and 212°F
Answer:
A sample of helium gas has a volume of 620mL at a temperature of 500 K. If we ... to 100 K while keeping the pressure constant, what will the new volume be?
Explanation: