The type of the bond is present Na₃PO₄ is the ionic bond. the Na₃PO₄ is the ionic compound. yes the Na₃PO₄ is the polyatomic ion.
The Na₃PO₄ is Na⁺ and PO₄³⁻. the phosphorus is the non metal and the oxygen atom is the non metal. the non meta and non meta form the covalent or molecular bond. the bond between the PO₄³⁻ bond is the covalent bond but the overall present in the Na₃PO₄ is the ionic bond . the bons in between the Na⁺ and PO₄³⁻ is the the ionic bond. the PO₄³⁻ id the polyatomic ion .
The bond between the positively charged ion and the negatively charged ion are called as the ionic bond and the compound form is the ionic compound.
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Answer:
In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Answer:
(b) Both have the same number of valence electrons.
Step-by-step explanation:
We find the most striking chemical similarities between two Main Group elements when they are in the same Group of the Periodic Table.
Elements in the same Group have the same number of valence electrons.
(a) is <em>wrong</em>, because elements in the same group have <em>different masses</em>.
(c) is <em>wrong,</em> because atoms with the same number of protons belong to the s<em>ame element</em>.
(d) is wrong, because elements in the same Group must be in .
<em>different Periods.</em>
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42