Answer:
A:Boyle's Law or B:Charles's Law
Explanation:
Diamonds are composed of carbon in a tetrahedral lattice. That is option C.
<h3>What is a diamond?</h3>
A diamond of defined as an allotrope or one of the major forms of the element, carbon in nature.
These carbon atoms are arranged within the diamond in a face centered cubic tetrahedral lattice shape.
Therefore, Diamonds are composed of carbon in a tetrahedral lattice.
Learn more about carbon here:
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Zn = 28.15%
Cl = 30.53%
O = 41.32%
<h3>Further explanation</h3>
Given
Zn(CIO3)2 compound
Required
The % composition
Solution
Ar Zn = 65.38
Ar Cl = 35,453
Ar O = 15,999
MW Zn(CIO3)2 = 232.3
Zn = 65,38/232.3 x 100% = 28.15%
Cl = (2 x 35.453) / 232.3 x 100% = 30.53%
O = (6 x 15.999) / 232.3 x 100% = 41.32%
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
Answer:

Explanation:
Here, we want to calculate the number of formula units in the given molecule
We start by getting the number of moles
To get the number of moles, we have to divide the mass given by the molar mass
The molar mass is the mass per mole
The molar mass of calcium bromide is 200 g/mol
Thus, we have the number of moles as follows:

The number of formula units in a mole is:

The number of formula units in 0.2075 mole will be: