Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH
[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
Answer:
Training officers in how to properly collect evidence
Explanation:
Forensic science is an interesting branch of science that involves the use of scientific procedures to solve a crime case. It encompasses collection of physical evidence from the crime scene and analyzing it in a laboratory using scientific means.
A forensic scientist is the individual in charge of performing these scientific procedures. His/her major role is to run the scientific analysis of the physical evidence brought in by the officers, however, he/she can also perform the task of training officers in how to properly collect evidence, in order not to damage the evidence or render it invalid for use.
Answer:
umm.. B. a base that generates a lot of hydroxide ions in water.
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
We are given that the balanced chemical reaction is:
cacl2⋅2h2o(aq) +
k2c2o4⋅h2o(aq) --->
cac2o4⋅h2o(s) +
2kcl(aq) + 2h2o(l)
We known that
the product was oven dried, therefore the mass of 0.333 g pertains only to that
of the substance cac2o4⋅h2o(s). So what we will do first is to convert this
into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is
molar mass of cac2o4 plus the
molar mass of h2o.
molar mass cac2o4⋅h2o(s) = 128.10
+ 18 = 146.10 g /mole
moles cac2o4⋅h2o(s) =
0.333 / 146.10 = 2.28 x 10^-3 moles
Looking at
the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is
1:1, therefore:
moles k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles
Converting
this to mass:
mass k2c2o4⋅h2o(aq) = 2.28
x 10^-3 moles (184.24 g /mol) = 0.419931006 g
Therefore:
The mass of k2c2o4⋅<span>h2o(aq) in
the salt mixture is about 0.420 g</span>
