Answer:

Explanation:
Given:
A solution contains one or more of the following ions such as Ag,
and 
Here the Lithium bromide is added to the solution and no precipitate forms
Solution:
Since with LiBr no precipitation takes place therefore Ag+ is absent
Here on adding
to it precipitation takes place.
Precipitate is as follows,

Thus,
is present
When
is added again precipitation takes place.
Therefore the reaction is as follows,

Therefore,
are present in the solution
Answer:
4.16g of MgCl2
Explanation:
First, let us generate a balanced equation for the reaction:
Mg + 2HCl —> MgCl2 + H2
Molar Mass of Mg = 24g/mol
Molar Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g
From the equation,
24g of Mg produced 95g of MgCl2.
Therefore, 1.05g of Mg will produce = (1.05x95)/24 = 4.16g of MgCl2
Answer:
See below ~
Explanation:
The calculated values of V/n :
⇒ 1.5/0.3 = 5
⇒ 3/0.6 = 5
⇒ 4.5/0.9 = 5
⇒ 6/1.2 = 5
⇒ 7.5/1.5 = 5
1. From this we understand that the calculated values of V/n remain constant, equal to <u>5</u> in this case.
2. The volume-mole graph will be a straight line passing through the origin. (Attached below)
I believe it is D. because stoichiometry allows people to determine how much reactant is consumed in a chemical reaction.
Answer:
See images attached and explanation
Explanation:
I have drawn three possible structures of ZX2. We have to remember that the shapes of molecules could be predicted on the basis of the Valence shell electron pair repulsion theory.
The number of electrons on the valence shell of the central atom determines the shape of the molecule. We have also been told that X is not hydrogen.
If the two X atoms are arranged at a bond angle of 180 degrees, we could have either structure I or II. We will have these structures if the Z atom is sp2 hybridized.
Similarly, if the Z atom is sp3 hybridized, we may have structure III in which the molecule is bent with a bond angle less than 109 degrees. This may result from the presence of a lone pair on Z.
Note that all these structures obey the octet rule.