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Aloiza [94]
2 years ago
14

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ulate the molarity of the solution.
Chemistry
1 answer:
ladessa [460]2 years ago
4 0

Answer:

The molarity of the solution is 1.1 \frac{moles}{liter}

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}

In this case

  • number of moles of solute= 0.564 moles
  • volume= 0.510 L

Replacing:

molarity=\frac{0.564 moles}{0.510 L}

Solving:

molarity= 1.1 \frac{moles}{liter}

<u><em>The molarity of the solution is 1.1 </em></u>\frac{moles}{liter}<u><em></em></u>

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<h2 /><h2 /><h2>answer.</h2>

five oxygen molecules

step by step explanation.

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Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p
masha68 [24]

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, (Br_2) = 159.8 g/mol

Putting values in above equation, we get:

\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol

Hence, the amount of liquid bromine produced is 4.79 moles.

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A ground state hydrogen atom absorbs a photon of light having a wavelength of 92.05 nm. It then gives off a photon having a wave
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Absorbed photon energy
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Ea = hc / 92.05^-9m 

<span>Energy emitted
Ee = hc/ 1736^-9m </span>

Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
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<span>Converting J to eV (1.60^-19 J/eV)
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∆E = 12.70 eV </span>

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<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>

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