Answer:
All atoms heavier than barium
Explanation:
In the periodic table, elements are divided into blocks. We have the;
s- block elements
p- block elements
d- block elements
f- block elements
However, immediately after Barium, we now encounter elements that have f-orbitals. Barium possesses a fully filled d-orbital. Hence after it, we see elements with 4f and 5f orbitals called the Lanthanides and actinides. The elements following the lanthanide and actinide series possess completely filled f-orbitals as inner orbitals.
Hence elements heavier than barium all possess f-orbitals.
I believe the answer is A the 1st one
3Na2O(at) + 2Al(NO3)3(aq) —> 6NaNO3(aq) + Al2O3(s)
This is a double replacement reaction and NaNO3 is aqueous because Na is an alkali metal, plus nitrate is in the solution. Both of these are soluble. Al2O3 is not soluble because it does not contain any element that is soluble and is hence the precipitate.
Hope this helped!
The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
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Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
Answer:
E°(Ag⁺/Fe°) = 0.836 volt
Explanation:
3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt
Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt
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Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...
E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt
