Al has a charge of +3. Cl has a charge of -1. Hence 3 chloride ions are required to neutralize charge of +3 on aluminium and the compound formed is $ AlCl_3$.
And of course, in an ionic solid (say $"sodium chloride"$), an isolated cation, a sodium ion, is BOUND ELECTROSTATICALLY to EVERY OTHER CHLORIDE ANION in the lattice. The attraction/repulsion follows...
Since the formula for potassium chloride is $"KCl"$, add the atomic masses of potassium and chlorine to determine the atomic mass of $"KCl"$. $"Atomic mass KCl"="39.1 u + 35.45 u"="74.6...
A completely filled outer electron shell is stable because it can not gain any more electrons, it is full. Being completely full it has no desire to lose electrons thereby...
Aluminum has a common oxidation state of $"3+"$, and that of iodine is $"1-"$. So, three iodides can bond with one aluminum. You get $"AlI"_3$. For similar reasons, aluminum chloride...
and bromine forms a $Br^(-)$ ion........... And so, if these ions are to make music together, and form a NEUTRAL ionic solid, we need 2 equiv of anion, and 1...
So $4.50$ mol $AlCl_3harr3xx4.50=13.50$ mol $Cl^-$ 1 mol contains $~~6.02xx10^23$ particles, so the answer is: $n_(Cl^-)=13.50xx6.02xx10^23=81.27xx10^23~~8.13xx10^24$ (rounded to 3 significant digit because of the $4.50$)
First you need to write the correct BALANCED chemical reaction equation. Then you can determine how many moles of $Cl_2$ would be produced from the available moles of Cl in...
1 mole of aluminum oxide has a molecular weight of 102 grams (Al=27 grams and O=16 grams). It means that for 1000 kg of compound you can get 529 kilograms...
We're asked to find the number of grams of $"Al"$ that reacted, given some $"H"_2 (g)$ product characteristics. Let's first write the chemical equation for this reaction: $2"Al"(s) +...
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