Use to look up the atomic masses of the in the compound. Then multiply the for each element times its subscript, and add them together.
Formula Mass of
For more information about atomic mass units and unified mass units, see the following link:
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 viewsGiven: $Ca(ClO_3)_2$ The compound is Calcium chlorate. The Parentheses is a multiplier for all of the inside the parentheses. $"Element Symbol No. of Atoms"$ Calcium $" "Ca: "...
1 Answers 1 viewsAnd $"sodium chlorate(V)"$ is the old sodium chlorate, $NaClO_3$. $ClO_4^-$; i.e. $Cl(+VII)$, $"perchlorate."$ $ClO_3^-$; i.e. $Cl(+V)$, $"chlorate."$ $ClO_2^-$; i.e. $Cl(+III)$, $"chlorite."$ $ClO^-$; i.e. $Cl(+I)$, $"hypochlorite."$ $Cl^-$; i.e. $Cl(-I)$, $"chloride."$ And...
1 Answers 1 viewsNormally, this rxn must be calalyzed with a little $Mn(IV)$ salt, but we address the question with this . We spit out a $19.7*mol$ quantity of dioxygen gas. Given the...
1 Answers 1 viewsStart by writing a balanced chemical equation for this $"Pb"("ClO"_3)_text(2(aq]) + color(red)(2)"NaI"_text((aq]) -> "PbI"_text(2(s]) darr + 2"NaClO"_text(3(aq])$ Lead(II) chlorate will react with sodium iodide to form lead(II)...
1 Answers 1 views$"Oxidation:"$ $stackrel(I^+)ClO^(-) + 2H_2Orarr stackrel(V^+)ClO_3^(-)+4H^+ +4e^(-)$ $(i)$ $"Reduction:"$ $stackrel(I^+)ClO^(-) + 2H^+ + 2e^(-) rarr stackrel(""^(-)I)Cl^(-)+H_2O$ $(ii)$ And so we takes....$(i)+2xx(ii)$ $stackrel(I^+)ClO^(-) +2stackrel(I^+)ClO^(-) cancel(+2H_2O+ 4H^+ + 4e^(-))rarr stackrel(V^+)ClO_3^(-)+2stackrel(""^(-)I)Cl^(-)cancel(+4H^+ +2H_2O+4e^(-))$ ...to give....after...
1 Answers 1 viewsA mole is defined as the amount of a substance whose mass in grams equals the atomic, molecular, or formula-unit mass in units. Example 1: Water ($"H"_2"O"$) has two hydrogen...
1 Answers 1 viewsWhile I have given the stoichiometric equation for the decomposition of the metal chlorate, this reaction is simpler than it seems. We know that a mass of $5.837-1.459=4.378$ $g$ was...
1 Answers 1 viewsYou'd need 33.3 moles of potassium chlorate, $KClO_3$, to produce that much oxygen. All you need in order to answer this question is the balanced chemical equation for the of...
1 Answers 1 views$"% yield"$ $=$ $"Equivalence mass of product"/"Equivalent mass of reactant"xx100%$. You have a stoichiometric eqaution: $Be(s) + 2HCl(aq) rarr BeCl_2(aq) + H_2(g)$. And you have a mass of product....
1 Answers 1 views