Answer:
Fewer bubbles will be produced because of fewer collisions of reactant molecules
Explanation:
As the solid dissolves into the solution after the liquid has been vigorously bubbled, if the temperature of the liquid is reduced a little, what will happen is that fewer bubbles will be produced as a result of lesser amount of collisions occurring between the reactant molecules
Answer:
Option A and D are correct.
Unstable species react rapidly.
Stable species do not react rapidly.
Explanation:
The complete question is attached to this solution.
The more stable a reactant is, the less reactive it will be. A stable reactant has a very stable structure in which it will avoid any perturbations. And for a reaction to occur, the bonds in the reactant must break down to form the products. A stable reactant has very strong bonds that aren't easy to break down, hence, reactions involving very stable reactants do not proceed rapidly.
And the more unstable a reactant specie is, the more rapidly it reacts. This is why the reaction involving the less stable isotope of carbon; Carbon-14 is very rapid. It is the same reason as explained above that is responsible for this. The bond between unstable species are not strong and are easily breakable, thereby leading to a quick reaction.
Hope this Helps!!!
Answer:
-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.
-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.
With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.
Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
Answer:
1 - 3
Explanation:
- Look to see where metals are on the periodic table then look at what group they are in. The group tells you the number of valence electrons. Ex. 1A has one valence electron.
- Hope this helped! If you need a further explanation please let me know.