This threefold difference in molar mass, of course translates to . Given that the difference in densities is (very approx.) threefold as well, we would predict that the atomic packing in both metals is similar in terms of , $"mass per unit volume"$. Does this make sense?
Volume of this cube $V= a^3$ $V=4.4^3$ $V=85.184 cm^3$ Since (d) is 21.4 g per cubic centimeter Mass will be $M=d*V$ $M=21.4*85.184 = 1822.94 grams = 1.82 kg$
$"Average atomic mass"$ is the weighted average of the individual isotopes. And thus the weighted average $=$ $(0.6917xx63+0.3083xx65)*"amu"$ $=$ $"?? amu"$. Note that we have not accounted for the...
$"% of platinum"$ $=$ $"Total mass of platinum"/"Total mass of alloy"xx100%.$ $(2%xx11*kg+36%xx6*kg)/(11*kg+6*kg)xx100%$ $=(2.38*kg)/(17*kg)xx100%=??%$ The spot price for platinum is currently $£25-24*g^-1$. What is this lump worth?
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...
One mole of an element contains $6.022xx10^23$ atoms of the element. $"1 mol Li"=6.022xx10^23"atoms Li"$. In order to convert atoms to moles, divide the given number of atoms by $6.022xx10^23"atoms/mol"$....
It is a fact that if there is a mass of approximately 1.01 g of hydrogen atoms, there are Avogadro's number of hydrogen ATOMS. It is also a fact that...
From your Periodic Table we learn that one mole of copper, $6.022xx10^23$ individual copper atoms have a mass of $63.55*g$. And thus we use the MASS of a chemical sample...
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...