Answer:
In analytical chemistry, the titrant is a solution of known concentration that is added (titrated) to another solution to determine the concentration of a second chemical species. The titrant may also be called the titrator, the reagent, or the standard solution.
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# Cary on learning
<span>the balanced equation for the reaction is as follows ;
Na</span>₂S + 2AgNO₃ ---> 2NaNO₃ + Ag₂<span>S
stoichiometry of Na</span>₂S to AgNO₃<span> is 1:2
number of AgNO</span>₃<span> moles reacted - 0.315 mol/L x 0.04000L = 0.0126 mol according to molar ratio of 1:2
number of Na</span>₂S moles required are - 1/2 x number of AgNO3 moles reacted Na₂<span>S moles = 0.0126 mol /2 = 0.00630 mol
molarity of Na</span>₂<span>S - 0.260 M
there are 0.260 mol in 1 L
therefore 0.00630 mol are in - 0.00630 mol / 0.260 mol/L
volume of Na</span>₂<span>S required = 0.0242 L
volume of Na</span>₂S required = 24.4 mL
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
Density is the mass of compound divided by its volume can be shown as follows:
40 mL of snow having 20 g of mass calculated from density.
Now, 10 cm of snow = 3.93 inches = 20 g
As, 10 inches of rain will produce 11 inches of ice as the volume of ice is bigger than rain water.
10 inches rain = 11 inches snow
3.93 inches of snow produced by
Thus, 3.57 incehs of rain produces by 10 cm snow.
<span>2NaCN + (1)H2SO4 → Na2SO4 + 2HCN
</span><span>The coefficient of sulfuric acid is 1.</span>