Answer:
substances with a higher boiling point are returning back to the flask which allows another substances with the specific context temperature (lower boiling point) to boil over and be purified.
Explanation:
The reason it happens because the lower boiling point substance vaporizes and crosses over while the other substance is waiting for its boiling point to reach
Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>
Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.
To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.
Explanation:
Equiv means equivalent ...
The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
The micromoles of mercury(II) iodide : 0.013 μ moles
<h3>Further explanation</h3>
Given
215.0mL of a 6.0x10⁻⁵mmol/L HgI₂
Required
micromoles of HgI₂
Solution
Molarity(M) = moles of solute per liters of solution
Can be formulated :
M = n : V
n = moles
V = volume of solution
V = 215 mL = 0.215 L
so moles of solution :
n = M x V
n = 6.10 mmol/L x 0.215 L
n = 1.312 . 10⁻⁵ mmol
mmol = 10³ micromol
so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles
