Copper oxide reacts with hydrochloric acid to produce copper chloride and water.
So the basic equation that illustrates this reaction is:
CuO + HCl ......> CuCl2 + H2O
Now, we need to balance this equation:
the number of copper and oxygen moles are balanced.
For each of hydrogen and chlorine, we have one mole as a reactant and two moles as product.
Therefore, to balance this, we will add an extra mole of HCl in the reactants.
Based on this, the final balanced equation would be:
CuO + 2HCl ................> CuCl2 + H2O
Each mole of a substance contains 6.02 x 10²³ particles
Atoms of Fe = 4.5 x 6.02 x 10²³
= 2.709 x 10²⁹ atoms
That's because <span>the specific heat capacity of water is higher than specific heat capacity of iron, meaning that the water would need to lose more heat energy to drop its temperature.</span>
a) Group 2 elements have 2 electrons on their outer shell, so they form a 2+ charge.
b) they lose 2 electrons as they are transferred to the non metal.
c)They obtain this charge as when they are made into an ionic compound the 2 electrons on the outer shell are transferred to the non metal, meaning there are 2 more protons that electrons, giving it a positive charge.
hope this helps! :)
Answer:
fraction of vacancies for this metal FV = 1.918*10⁻⁴
Explanation:
Given:
The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )
But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol
Density of the material is given in g/cm3 we need to convert it to g/m³
Density of material ( p ) in g/m³ :
To convert we know that
1 g/cm³ = 1000000 g/m³ then
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³
Given Atomic mass = 85.5 g/mol
To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
We can now Calculate the fraction of vacancies using the formula below;
Fv = Nv / N
Fv = 1*10²⁵ / 5.212*10²⁸
fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴
