Answer:
73.41 g
Explanation:

First find number of moles of Mg
n(Mg)
Second, find the number of moles of magnesium nitrate produced by this number of moles of magnesium.
The molar ratio is one is to one (using stoichiometric coefficients from the balanced equation). So 5.377 mol of magnesium produces 5.377 mol of magnesium nitrate
Third, find what mass of magnesium nitrate that is
mass(Mg(NO3)2) = mol * molar mass
= 5.377 mol * 148.3148 g/mol
= 148.3148 g
But the system is inefficient. It only has an efficiency of 49.50 so what will be the actual, not theoretical yield.
actual yield = efficiency * theoretical yield
= 49.5% * 148.3148 g
= 73.41 g
The element that has been oxidized is Al ( Answer A)
Explanation
if the oxidation number of an element is greater in the product side than the reactant side then the element has lost the electrons . This mean that the element has been oxidized.
Al is oxidized because it move from oxidation state Zero(0) in reactant side to oxidation state 3⁺ in product side.
This implies that Al loses 3 electrons.
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)


Given s = 7.4×10⁻³ M
So, Ksp is:


Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)

Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>