<span>3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
The unbalanced equation is:
ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s)
First, count the atoms of each element on each side of the equation:
Zn 1,1
Br 2,3
Al 1,1
The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting:
3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s)
Now check the atom counts again for both sides:
Zn 3,1
Br 6,6
Al 1,2
Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
Now check the atom counts again:
Zn 3,3
Br 6,6
Al 2,2
And they match. So the balanced equation is:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)</span>
Answer:
D
Explanation:
I - You should account the mass of the weighing paper to reduce it from the total mass at the end of the process, having only the mass of the Silver Chloride.
II - The precipitation of the silve chloride will occur independently of the temperature, because the Kps of this salt is very low (Ksp = [Ag+] .[Cl-]).
III - Washing the precipitate will secure the purity of the final product, it won't allow any other contaminant to be in your precipitante which could change your final mass.
IV - You should heat you AgCl precipitate so it will be dry, because of that the mass you will obtain is only the mass of the weighing paper and the silver chloride and nothing else.
The pair which consist of molecules having the same geometry is CH2CCI2 and CH2CH2.
Both of these molecules contain double bonds, which has sp^2 hybridization and they possess a trigonal planar geometry. In trigonal planar geometry, the molecule consist of three equally spaced sp^2 hybrid orbitals, which arranged at angle 120 degree.
Answer:
Here's what I get.
Explanation:
The frequency of a vibration depends on the strength of the bond (the force constant).
The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.
Acetophenone
Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.
p-Aminoacetophenone
The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.
The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.
p-Nitroacetophenone
The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.
As electron density moves toward C-1, the double bond character of the C=O group increases.
The bond length decreases, so the bond becomes stronger, and wavenumber increases to 1693 cm¹.
Answer:
I think that the trend that would be seen in the time column of the data table would be that the number of seconds would increase. I know this because for each flask, the concentration of sodium thiosulfate decreases, since less of it is being mixed with more water. Also, when the concentration of a substance decreases, then the reaction rate also decreases, as there will be fewer collisions with sulfuric acid if there are fewer moles of sodium thiosulfate. When there are fewer collisions in a reaction, the reaction itself will take longer, and so when the sodium thiosulfate is diluted, the reaction takes more time.
Explanation:
<em>I verify this is correct. </em>
