Answer:
1.27 x 10⁻⁴M
Explanation:
For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.
Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³
Solve for 'x' => x = Solubility = ∛Ksp/4
Answer:
1.3 × 10⁴ mL
Explanation:
A Pb-contaminated water sample contains 0.0012 %Pb by mass, that is, there are 0.0012 g of Pb in 100 g of solution. The mass of the sample that contains 150 mg (0.150 g) of Pb is:
0.150 g Pb × (100 g sample/0.0012 g Pb) = 1.25 × 10⁴ g sample
The density of the sample is 1.0 g/mL. The volume of the sample is:
1.25 × 10⁴ g × (1 mL/1.0 g) = 1.3 × 10⁴ mL
The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:
-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)
where C is the specific heat capacities of the materials.
We calculate as follows:
-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C<span>
</span>
Atomic number of Na -11
electronic configuration of Na - 2,8,1
outermost shell of Na has one electron. To become stable and to have a complete octet in the outermost shell, it loses that electron in the outermost shell and becomes positively charged.
Then electronic configuration is 2,8 with 10 electrons . Noble gas with atomic number 10 is Neon.Therefore it will have the same number of valence electrons as B) Neon
