Answer:
0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.
Explanation:
This is a problem of dilution using the equation:
<em>initial concentration x initial volume = final concentration x final volume.</em>
The final volume to be prepared is 25 microliters.
The final concentration to be prepared is 3 M.
The initial volume to be taken is not known yet.
The initial concentration is 10 M.
Now, let's substitute these parameters into the the equation above.
10 x initial volume = 3 x 25
Initial volume = 3 x 25/10
= 7.5 microliters
Note that: 1 microliter = 0.001 milliliters
Hence,
7.5 microliters = 0.0075 milliliters
<u>This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.</u>
The mass of NaN3 needed to produce 17.2 L nitrogen at STP is calculated as follows
find the moles of N2 produced at STP
At STP 1mole of gas = 22.4 L , what about 17.2 L of nitrogen
by cross multiplication
= (1 mole x17.2 L)/ 22.4 L= 0.768 moles
2NaN3 =2Na +3 N2
by use of mole ratio between NaN3 to N2 (2:3) the moles of NaN3 = 0.768 x2/3 = 0.512 moles of NaN3
mass of NaN3 is therefore =moles of NaN3 xmolar mass of NaN3
=0.512moles x 65 g/mol =33.28 grams of NaN3
Did you ever get the awnser to this question cause im stuck on it too
