Answer:
- NH₄CN,
- NH₄ClO₃,
- (NH₄)₂SO₃,
- NH₄H₂PO₄.
- Al(CN)₃,
- Al(ClO₃)₃,
- Al₃(SO₃)₃,
- Al(H₂PO₄)₃.
Explanation:
Cations (positive ions) and anions (negative ions) combine to form ionic compounds. The charge of each ion is stated as the superscript in its formula.
The final compounds should be neutral (not carrying charge.) For that to happen, the ions should combine at a ratio that allows their charges to balance.
By convention, in chemical formula of ionic compounds, cations are written before anions.
<h3>Example: Al³⁺ and SO₃²⁻</h3>
The charge on an Al³⁺ cation is +3. The charge on a SO₃²⁻ anion is -2.
Start by finding the least common multiple of the absolute value of the charges (drop the negative sign in front of the charge.)
.
The number of Al³⁺ ions in a formula unit of this compound will be equal to
.
Similarly, the number of SO₃²⁻ ions in a formula unit of this compound will be equal to
.
If a polyatomic ion (an ion with more than one atoms in it) appears more than once in the formula, enclose it in brackets and write the number of its occurrence outside the bracket.
.
Apply this rule to find the formula of other compounds. Additionally, keep in mind that if the charge on the cation and the anion is the same, each should occur only once in the formula.
Answer:
2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom
Explanation:
Given electronic configurations are :
1st:
2nd :
given 1st ionization energy are: 2080 kJ/mol and 496 kJ/mol
generally ionization energy of fulfilled orbital is more than half filled orbital and these two state are more stable.
therefore ionization energy of fulfilled is more than half filled orbital
hence
ionization energy of 1st atom will be very high because its orbital is fulfilled and less energy for 2nd atom so 2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom.
They contain both carbon and hydrogen :)
Answer:
The percent of mass of GaBr₃ in the solid mixture is 30.2 %.
Explanation:
GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)
MW GaBr₃ = 309.4 g/mol
MW AgBr = 187.8 g/mol
187.8 g AgBr _______ 1 mol
0.368 g AgBr _______ x
x = 2.0 x 10⁻³ mol AgBr
1 mol GaBr₃ ____ 3 mol AgBr
y ____ 2.0 x 10⁻³ mol AgBr
y = 6.7 x 10⁻⁴ mol GaBr₃
1 mol GaBr₃ ____________ 309.4 g
6.7 x 10⁻⁴ mol GaBr₃ ______ w
w = 0.206 g GaBr₃
0.6813 g _____ 100%
0.206 g _____ z
z = 30.2 %