Do you agree? The copper displaces the given volume of water. Now $rho_"Cu"=8.90*g*cm^3$ OR $rho_"Cu"=8.90*g*mL^-1$, i.e. $1*mL-=1*cm^3$ But by definition, $rho_"density"="mass"/"volume"$ And thus $"mass"=rhoxx"volume"=8.90*g*cancel(mL^-1)xx23.4*cancel(mL)$ $=208.3*g$. Do you follow?
1 Answers 1 viewsYou have given the stoichiometric equation that represents reduction of cuprous sulfide: $Cu_2S(s) + O_2(g) rarr 2Cu(s) + SO_2(g)$ Sulfur is oxidized; copper and oxygen are reduced. $"Moles of copper"=(650*g)/(63.55*g*mol^-1)=10.3*mol$....
1 Answers 1 viewsThis reaction is balanced with respect to mass and charge; as indeed it must be if it reflects chemical reality. Most carbonates undergo this decomposition, provided you supply enuff heat...
1 Answers 1 viewsBased on the information I got from you: $m_"Zn" = "6.530 g"$ $m_("CuSO"_4cdot5"H"_2"O") = "5.00 g"$ (in lab you tend to be handed the hydrate) From the molar...
1 Answers 1 viewsThe idea here is that you need to use the mass of copper and the mass of the copper sulfide to determine how much sulfur the produced compound contains, then...
1 Answers 1 viewsThe empirical formulas are $"Cu"_2"O"$ and $"CuO"$. Oxide 1 $color(white)(mmmmml)"Cu" +color(white)(m) "O" → "Oxide 1"$ $"Mass/g":color(white)(l) 2.118color(white)(ll) 0.2666$ Our job is to calculate the ratio of the moles of...
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 viewswill generally dissociate into ions in solution. For the purposes of this problem, let's assume there isn't much ion pairing. The key here is knowing we're worried about...
1 Answers 1 viewsWe know that the sea water contains $(65xx10^-3"g of bromide ions")/L $ If you want $Br_2$ the following reaction takes place $Br^-) + Br^-) = Br_2 + 2e^-$ Well this...
1 Answers 1 viewsThe Lewis concept of acidity designates acids as electron pair acceptors, and bases as electron pair donors. For the reaction: $AlBr_3 + Br^(-) rarr AlBr_4^-$, clearly the bromide ion is...
1 Answers 1 views