Use a magnet to separate the iron from the sand.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.
There are 1.92 × 10^23 atoms Mo in the cylinder.
<em>Step 1</em>. Calculate the <em>mass of the cylinder
</em>
Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g
<em>Step 2</em>. Calculate the<em> mass of Mo
</em>
Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo
<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo
</em>
Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo
<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo
</em>
Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)
= 1.92 × 10^23 atoms Mo
The answer is B. A mixture can be separated as shown in the example.<span />
