moles Cu produced : 0.002
<h3>Further explanation</h3>
Concentration of copper sulfate (CuSO₄) : 0.319 g/dm³
MW CuSO₄ :

mol CuSO₄ /dm³ :

CuSO₄⇒Cu²⁺ + SO₄²⁻
mol Cu : mol CuSO₄ = 1 : 1 , so mol Cu²⁺=0.002
9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

The mass of nitrogen in the second sample is 23.33 g.
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre