So, you know that your atom contains $16$ protons $18$ neutrons $16$ electrons Right from the start, you know that you're indeed dealing with a neutral...
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1 Answers 1 viewsThe weighted average is the sum of each individual nuclide, multiplied by its isotopic abundance. So, we simply do the arithmetic: $0.7577xx34.969+0.2423xx36.966 = ???$ amu Then look at to...
1 Answers 1 viewsAtomic radii DECREASE across a Period, but increase down a Group. Why? As we add another proton to the nucleus we add another electron to the valence shell. As we...
1 Answers 1 viewsIt wants the configuration that helium has, which is two . To be more specific, it wants a DUET, not an octet. Beryllium ($"Be"$) has the $4$ (not $2$),...
1 Answers 1 viewsPhosphorus has $15$. As such its electronic configuration in ground state is $1s^2 2s^2 2p^6 3s^2 3p^3$ The following electronic configurations could be excited states $1s^2 2s^2...
1 Answers 1 viewsIf the sodium is ionised, the outer $3s^1$ electron is lost, so the configuration becomes $1s^2 2s^2 2p^6$, which is the same as neon, $Ne$.
1 Answers 1 viewsFor starters, you know that your element is located in period $4$ of because all the transition metals that have $3d$ configurations are located in period $4$ of the Periodic...
1 Answers 1 viewsYou have filled 2 shells, and begun a third valence shell. Is this all you want, or have I missed the point?
1 Answers 1 views$MgCO_3(s)+DeltararrMgO(s) + CO_2(g)$ And we would have to heat the magnesium salt VERY fiercely to get complete decarboxylation. $"Moles of carbon dioxide LOST"$ $=$ $(4.40*g)/(44.01*g*mol^-1)$ $=0.010*mol$ And therefore in...
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