The 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 viewsYou don't need to note charges in the equation, if that's what you meant. The charge is implied because of the three Chlorides. Hope this helps!
1 Answers 1 viewsAluminum 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...
1 Answers 1 viewsUnbalanced Equation $"Cl"_2" + KBr"$$rarr$$"KCl" + "Br"_2"$ Balance the Cl There are 2 Cl atoms on the left side and 1 Cl atom on the right side. Add a coefficient...
1 Answers 1 viewsIn , the must add together to become $0$. In aluminum bromide, aluminum has an oxidation number of $+3$ and bromine has an oxidation number of $-1$. In order...
1 Answers 1 views$"Moles of arsenic"$ $=$ $(1.587*g)/(74.92*g*mol^-1)=0.0212*mol$. $"Moles of chlorine"$ $=$ $(3.755*g)/(35.45*g*mol^-1)=0.106*mol$. We divide thru by the lowest molar quantity to give $AsCl_5$. This is not chemical you would find in a...
1 Answers 1 viewsFirst 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 Answers 1 viewsEach equiv of metal reduces $3$ equiv of acid, and $3/2$ equiv of dihydrogen gas are evolved. $"Moles of aluminum"$ $=$ $(0.498*g)/(26.98*g*mol^-1)=0.0185*mol$. Now, given the conditions, clearly the metal is...
1 Answers 1 viewsGiven: Volume of $"H"_2$ at STP; $P$ and $T$ of $"HCl"$; chemical equation (understood) Find: Volume of $"HCl"$ Strategy: The central part of any problem is to convert moles...
1 Answers 1 viewsCalcium bromide speciates in aqueous solution according to the following equation......... $CaBr_2(s) stackrel(H_2O)rarrCa^(2+) + 2Br^-$ And thus for every metal ion there are necessarily two bromide ions. Here by...
1 Answers 1 views