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3 years ago

Find the original concentration of a solution diluted from 15.00 mL to 125.0 mL with a final concentration of 0.00383 M

Chemistry

1 answer:

BabaBlast [244]3 years ago

8 0

<h3>Answer:</h3>

0.03192 M

<h3>Explanation:</h3>

We are given;

Initial volume before dilution, V1 = 15.0 mL or 0.015 L
Volume after dilution, V2 = 125.0 mL or 0.125 L
Concentration after dilution, M2 = 0.00383 M

We are required to calculate the initial concentration, M1 before dilution.

To answer the question, we will use the dilution formula;

Dilution formula is given by; M1V1 = M2V2

Rearranging the formula;

M1 = M2V2 ÷ V1

= ( 0.00383 M × 0.125 L) ÷ 0.015 L

= 0.03192 M

Therefore, the original concentration of the solution before dilution is 0.03192 M

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

<u>Answer:</u> The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago

What is the formula mass/molar mass of H2CO2

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H - 1.01

C - 12.01

0 - 16.00

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3 years ago

4. If you put 1 mL of coffee into 9 mL of water, how much was the coffee diluted?

garri49 [273]

1 in 10, or 1/10, or 0.1 dilution factor It may be wrong but this is what I have!

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3 years ago

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steposvetlana [31]

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3 years ago

In one compound of nitrogen and oxygen, 0.615 grams of nitrogen combines with 0.703 grams of oxygen. In another 1.27 grams of ni

Orlov [11]

Answer:

Answer in explanation

Explanation:

In the first case, we divide each of the masses by the respective atomic masses:

N =0.615/14 = 0.043928571428571

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It can be seen here that the values are similar, hence the formula is NO

now let us look at the second data set:

N = 1.27/14 = 0.090714285714286

O = 2.9/16 = 0.18125

We now divide by the smallest

N = 090714285714286/090714285714286 = 1

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The formula here is thus NO2.

It can be seen that there are different oxides of nitrogen here which clearly indicates the law of multiple proportion.

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4 years ago