This question is basically asking us to convert the volume from mL to kL.
We can do this by setting up a multiplication problem and cancelling out units:
So now we know that
the Bay water will occupy 1.25x10^-4 kL.
Position of element in periodic table is depend on the electronic configuration of element.
Element with 62 electrons has following electronic configuration:
<span>1s2 2s2 </span>2p6 <span>3s2 </span>3p6 4s2 3d10 4p6 <span>5s2 </span>4d10 5p6 4f6 <span>6s<span>2
</span></span>
From above electronic configuration, it can be seen that highest value of principal quantum number, where electron is present, is 6. Hence, element belongs to 6th period.
Further, last electron has entered f-orbital, hence it is a f-block element. Position of f-block element is the bottom of periodic table.
Further, there are 6 electrons in f-orbital. Hence, it is the 6th f-block element in 6th period of periodic table.
Answer: 0.5 mole Mg
Explanation: solution:
12 g Mg x 1 mole Mg / 24 g Mg
= 0.5 mole Mg
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
