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Anarel [89]
1 year ago
10

If the absorbance of a KMnO4 solution of unknown concentration is 0.633, calculate the concentration of KMnO4 in the solution.

Chemistry
1 answer:
solong [7]1 year ago
3 0

The molar Concentration of KMnO₄ is 0.000219 M

Concentration is the abundance of a constituent divided by means of the overall extent of an aggregate. numerous styles of mathematical description may be outstanding: mass awareness, molar awareness, variety concentration, and quantity awareness.

y is absorbance

x is the molar concentration of KMnO_4

y = 4.84E + 03x - 2.26E - 01

0.833 = 4.84 * 10⁺⁰³ x - 2.26 * 10⁻¹

1.059 = 4.84 * 10⁺⁰³ x

X = 0.000219 M

Hence, The molar Concentration of KMnO₄ is 0.000219 M

Learn more about concentration here:-brainly.com/question/14469428

#SPJ9

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How many grams are in 1.76 x 10^23 atoms of iodine
Mariana [72]

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

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A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla
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Answer:

30 μmol/L

Explanation:

<em>A chemist prepares a solution of barium chloride by measuring out 8.9 μmol of barium chloride into a 300mL volumetric flask and filling the flask to the mark with water. </em>

<em> Calculate the concentration in μmol L of the chemist's barium chloride solution. Round your answer to 2 significant digits.</em>

<em />

The chemist has 8.9 μmol of solute (barium chloride) and he adds water until the mark of 300 mL in the container, which is the volume of the solution. We will need the conversion factor 1 L = 1000 mL. The concentration of barium chloride in μmol/L is:

\frac{8.9\mu mol}{300mL} .\frac{1000mL}{1L} =30\mu mol/L

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