This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Answer:
number of molecules =9.03×10^23
Explanation:
use the equation
n=number of molecules /Av.constant
make number of Molecules become the subject of the formula by multplying Av.constant on both sides of the equation,
number of molecules=n×Av. constant ,where Av.constant is 6.02×10^23
number of molecules=1.5mol×6.02×10^23
=9.03×10^23
Answer:
1.40 atm is the pressure for the gas
Explanation:
An easy problem to solve with the Ideal Gases Law:
P . V = n . R .T
T° = 370K
V = 17.3L
n = 0.8 mol
Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K
P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm
Its tetrahedral since there's four chlorine molecules around carbon
Answer:
The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing
we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633
In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.
a = 2r ..............(1)
other attached files show additional solutions in steps
