We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Answer:
Answer:
Number of Significant Figures: 5
The Significant Figures are 3 0 6 7 0
Explanation:
hope this helps
