So we could oxidize a single atom up to a multi-charged cation.........
$Cu+DeltararrCu^(29+) + 29e^-$
Of course we would need a LOT of energy to do this, and so the equation is (HIGHLY!) physically unreasonable. The ion still has $Z=29$ (manifestly), and this is the smallest such copper particle.
Typical oxidation states of copper are $Cu^(+I)$, and $Cu^(+II)$, $"cuprous"$ and $"cupric"$ ions.
The three basic principles underlying cell theory as we know it today are: All organisms are made of one or more cells. Cells are the basic building blocks of...
Most can exist as mono-atomic molecules, but some have molecules consisting of more than one atom of the element. Examples are $F_2,Cl_2, Br_2, O_2, N_2, P_4$ An element like sulphur...
If you change the atom by changing the number of protons you change the element to another element. Each element has unique properties based on the behavior of the...
And $"ductility"$ means that the material is capable of being drawn into a wire. And a second property that metallic , metal ions in an electron sea, confers is electrical...
You 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. $"Moles of copper"=(650*g)/(63.55*g*mol^-1)=10.3*mol$....
$CuCO_3(s) + 2HCl(aq)rarrCuCl_2(aq) + CO_2(g)uarr + H_2O(l)$ Or........... $CuO(s) + 2HCl(aq)rarrCuCl_2(aq) + H_2O(l)$ For each acid base reaction the product is the beautiful blue-coloured $[Cu(OH_2)_6]^(2+)$ ion, which is commonly represented...
$%A$ $=$ $(34.5*g)/(34.5*g+18.2*g+2.6*g)xx100%=62.3%$ $%B$ $=$ $(18.2*g)/(34.5*g+18.2*g+2.6*g)xx100%=33.0%$ $%C$ $=$ $(2.6*g)/(34.5*g+18.2*g+2.6*g)xx100%=4.70%$ That is $%" element"$ $=$ $"Mass of element"/"Mass of compound"xx100%$ $"Mass of compound"$ is clearly the SUM of the constituent ....
The 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...
The 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...