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LuckyWell [14K]
3 years ago
9

A solution of benzene is prepared by dissolving 0.32 mg of benzene to volume of 100 dm3 calculat molarity

Chemistry
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

4.10 x 10⁻⁸  M

Explanation:

The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. We have to first calculate the number of moles of solute (benzene) from the mass and the molecular weight:

Benzene has the chemical formula C₆H₆. So, its molecular weight (MW) is calculated from the molar mass of C (12 g/mol) and H (1 g/mol):

MW(C₆H₆)= (12 g/mol x 6 C) + (1 g/mol x 6 H) = 78 g/mol

Then we convert the mass of benzene (0.32 mg) from mg to grams and divide the mass into the MW:

mass = 0.32 mg x 1 g/1000 mg = 3.2 x 10⁻⁴ g

moles of benzene = mass/MW = 3.2 x 10⁻⁴ g/(78 g/mol) = 4.10 x 10⁻⁶ mol

Now, we convert the volume of solution from dm³ to L:

liters of solution = 100 dm³ x 1 L/1 dm³ = 100 L

Finally, we divide the moles of benzene into the volume of solution in liters:

M = moles of benzene/liters of solution = 4.10 x 10⁻⁶ mol/100 L = 4.10 x 10⁻⁸ mol/L = 4.10 x 10⁻⁸  M

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lana66690 [7]

Answer:

\text{ }1.25\times10^{23}\text{ formula units}

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:

\frac{41.5}{200}\text{ = 0.2075 mol}

The number of formula units in a mole is:

1\text{ mole = 6.02 }\times10^{22}\text{ formula units}

The number of formula units in 0.2075 mole will be:

0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}

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1 year ago
If you flow a solution of Mg CO3-- in water over a Cation resin, what will be in the outlet stream leaving the bed
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In a a cation-exchange resin, the outlet stream leaving the bed will contain H^{+} and CO_3 ^{2-}.

<h3>What is cation-exchange resin?</h3>
  • A resin or polymer that serves as a medium for ion exchange is known as an ion-exchange resin or cation-exchange resin.
  • It is an insoluble matrix (or support structure) made from an organic polymer substrate, typically appearing as tiny (0.25-1.43 mm radius) microbeads that are white or yellowish in color.
  • The process is known as cation-exchange resin because the beads are often porous, providing a wide surface area on and inside them where the trapping of ions takes place along with the concomitant release of other ions.
  • cation-exchange resin comes in many different varieties. Polystyrene sulfonate is the main ingredient in most commercial resins. Many diverse separation, purification, and decontamination techniques use cation-exchange resin.
  • The most typical examples are water filtration and water softening.

To learn more about cation-exchange resin with the given link

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What is an example of a microorganism acting in a harmful way? A) Bacteria aiding digestion. B) Bacteria causing strep throat. C
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B <<<< answer. Who would like to get strep throat. It's a fate for your worst enemy.

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4 years ago
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Is there any evidence that methanol forms ions either in the pure state or when dissolved in water?
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3 years ago
PLEASE HELP FAST! When 23.3 g of O2 reacts with 18.3 g C10H8, what is the limiting reactant. The equation is C10H8 + 12O2 --&gt;
Murljashka [212]

O₂ is the limiting reagent.

<h2>What is a limiting reactant?</h2>

A limiting reactant is described as the one that will be consumed first in a chemical reaction.

<h3>Calculation</h3>

C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O

Given, Mass of O₂ = 23.3 g

           Mass of C₁₀H₈ = 18.3 g

Molar mass of O₂ = 32 g

No. of moles of O₂ = 32/23.3 = 1.4 moles

Molar mass of C₁₀H₈ = 128 g

No. of moles of C₁₀H₈ = 128/18.3 = 6.9 = 7 moles

According to the equation,

1 mole of C₁₀H₈ yields 10 moles of CO₂. So, 7 moles of C₁₀H₈, give 70 moles of CO₂.

12 moles of O₂ yields 10 moles of CO₂. So, 1.4 moles of O₂ give 1.16 moles of CO₂.

We do know, however, that based on our balanced equation, if we were to totally react all 1.4 moles of O2, we could only produce a maximum of 1.16 moles of CO2. Even while there is enough C10H8 to generate more, the amount of O2 we have is a limiting factor because it will be consumed first.

So, our limiting reactant is O₂.

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