Answer:
Ag⁺(aq) + I⁻(aq) → AgI(s)
Explanation:
Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.
In the reaction:
AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).
The ionic equation is:
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).
Now, listing only the species that are participating in the reaction:
<h3>Ag⁺(aq) + I⁻(aq) → AgI(s)</h3>
The true statement should be B
the electron configuration for Phosphorous is 1s2 2s2 2p6 3s2 3p3
the diagram for 3p3 should be ^_ ^_ ^_ because you want to fill to put an electron in each slot before you double up
Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
elements are classified to study their properties and similarities and differences between each other more aqurately
hope it helps...!!!!
<h3>Answer:</h3>
Rubidium (Rb)
<h3>Explanation:</h3>
Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".
<h3>Trends in Periodic table:</h3>
Along Periods:
Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.
Along Groups:
Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.
<h3>Conclusion:</h3>
Given elements belong to same group hence, Rubidium present at the bottom of remaining elements will have least ionization energy due to facts explained in trends of groups above.
