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 viewsAnd 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...
1 Answers 1 viewsComparing their electronegativities will help determine the type of bond. is the tendency of a bonded atom to attract shared electrons to itself. The difference in electronegativity $(Delta"EN")$ is used...
1 Answers 1 viewsCommonly, ions are formed by (i) the loss of electrons from metals...electron rich materials from the left of as we face the Table, whose are furthest removed from the positively...
1 Answers 1 viewsBoth bonds result in forming of differentiating properties and structures. Atoms react in order to achieve stability. However, covalent bonds have a unique property that ionic bonds do not...
1 Answers 1 views$"Sodium chloride, potassium chloride, rubidium chloride,"$ $"caesium chloride,"$ all have a formula of $MCl$, where $M$ is the alkali metal. All the alkali metals have similar electronic configurations; thus they...
1 Answers 1 viewsWe know that $1*mol$ contains $6.022xx10^23$ individual sodium atoms, and has a mass of $22.99*g$. And thus for $1.$ there is a $2*g$ mass of $Na$. For $2.$ there is...
1 Answers 1 viewsThe idea here is that you need to use the given mass ratio to determine how much fluorine will be produced if the also produces $"34.5 g"$ of sodium....
1 Answers 1 views$"Moles of sodium chloride"$ $(581*cancelg)/(58.44*cancelg*mol^-1)$ $=$ $"approx. "10*mol$ Thus $10$ $mol$ sodium metal are required.
1 Answers 1 viewsYou know that $2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$ and that the reaction produced $"234 g"$ of sodium chloride. Convert this to moles by using the...
1 Answers 1 views