Answer:
Explanation:
Step 1. Identify the Group that contains X
We look at the consecutive ionization energies and hunt for a big jump between them
We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.
We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.
So, X is in Group 2 and P is in Group 15.
Step 2. Identify the Compound
X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.
We must have 3 X atoms for every 2 P atoms.
The formula of the compound is 
.
Answer:
Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃
The coefficients are: 1, 6, 3, 2
Explanation:
__Fe₂(SO₄)₃ + __KOH —> __K₂SO₄ + __Fe(OH)₃
To determine the correct coefficients, we shall balance the equation. This can be obtained as follow:
Fe₂(SO₄)₃ + KOH —> K₂SO₄ + Fe(OH)₃
There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by writing 2 before Fe(OH)₃ as shown below:
Fe₂(SO₄)₃ + KOH —> K₂SO₄ + 2Fe(OH)₃
There are 6 atoms of OH on the right side and 1 atom on the left side. It can be balance by writing 6 before KOH as shown below:
Fe₂(SO₄)₃ + 6KOH —> K₂SO₄ + 2Fe(OH)₃
There are 6 atoms of K on the left side and 2 atoms on the right side. It can be balance by writing 3 before K₂SO₄ as shown below:
Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃
Now, the equation is balanced.
Therefore, the coefficients are: 1, 6, 3, 2
Molar mass:
Ag = 107.86 g/mol
Cu = 63.54 g/mol
Mole ratio:
Cu(s)+2 AgNO₃(aq)→Cu(NO₃)₂(aq)+2 Ag(s)
63.54 g Cu ------------- 2 x 107.86 g Ag
1.50 g Cu -------------- ??
Mass Ag = 1.50 x 2 x 107.86 / 63.54
Mass Ag = 32358 / 63.54
= 509.25 g of Ag
hope this helps!
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