CuCl2 is Dichloride Copper, or Copper II Chloride.
In this Molecule, there is 1 atom of Copper for every 2 atoms of Chlorine.
If you go to the periodic table, you'll see that copper has a mass of 63.5 amu, and the chlorine 35.45 amu.
Which mean that the mass of CuCl2 is 63.5 + 35.45*2 = 63.5 + 70.9 = 134.4 g/mol.
Then, you find the ration of the mass of each atom and multiply by 100.
Ratio of Copper:
and you get about 47.2%
Ration of chlorice (Cl2):
and you get about 52.8%.
So, the percent composition of CuCl2 is 47.2% of Copper and 52.8% of chlorine.
Hope this Helps :)
Answer:
A strong base, such as NaOH. The amount of OH added shouldn't exceed 0.35 mols (though i would stop at 0.30 mols)
Explanation:
a weakly basic salt can be turned into a buffer by the addition of a strong base, and a weakly acidic salt can be turned into a buffer with a strong acid
Answer:
Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode
Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode
Explanation:
We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.
The overall reaction equation is;
2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)
E°cell= E°cathode - E°anode
E°cathode= 1.08 V
E°anode= 0.15V
E°cell = 1.08-0.15 = 0.93 V
But
∆G°= -nFE°cell
n= 2, F=96500C, E°cell= 0.93V
∆G° = -(2× 96500× 0.93)
∆G= -179490 J
But;
∆G = -RTlnK
R=8.314 JK-1
T= 25+273= 298K
Kc= the unknown
∆G° = -179490 J
Substituting values and making lnK the subject of the formula
lnK= ∆G/-RT
lnK= -( -179490/8.314 × 298)
lnK= 72.45
K= e^72.45
K= 2.91×10^31
Answer:
Add water to 29.22 grams of salt until you get 1 liter of volume, stir it until the salt completely dissolves and mixes and voila! you will have a 0.5 M NaCl solution.
:)
<span>A fast moving stream of air has a lower air pressure than a
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air pressure underneath the paper stayed the same. The
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</span>The
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the still air forces the balloon back into the lower pressure
of the air stream. Bernoulli’s Principle at work again!
