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omeli [17]
3 years ago
15

Using the following equation:

Chemistry
1 answer:
Lemur [1.5K]3 years ago
4 0
I think this is it...

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If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution
Helen [10]

This is a straightforward dilution calculation that can be done using the equation

M_1V_1=M_2V_2

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:

M_2=\frac{M_1V_1}{V_2}.

Substituting in our values, we get

\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].

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.

5 0
3 years ago
How many HCL-molecules in 1,5mol HCL
Vera_Pavlovna [14]

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

3 0
3 years ago
How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space
dezoksy [38]

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

7 0
3 years ago
Read 2 more answers
When these bond to make ccl4, what is the shape of a ccl4 molecule? linear circular tetrahedral trigonal planar
Aleks04 [339]
Its tetrahedral since there's four chlorine molecules around carbon
4 0
4 years ago
Read 2 more answers
Show that the ideal c/a ratio (height of unit cell divided by edge length) for the HCP unit cell is 1.633. (You may wish to refe
luda_lava [24]

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

7 0
3 years ago
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