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algol13
3 years ago
9

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr

opane concentration is 0.0445 m, what will the cyclopropane concentration be after 235.0 min
Chemistry
1 answer:
sergey [27]3 years ago
6 0

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices
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Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that occupies.
  • T is its temperature.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

  • P= 0.980 arm
  • V= 1.20 L
  • T= 287 K
  • R= 0.082 \frac{atmL}{molK}
  • n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082\frac{atmL}{molK}× 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082\frac{atmL}{molK}× 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

molar mass=\frac{0.458 grams}{0.04997 moles}

Solving:

<u><em>molar mass= 9.17 </em></u>\frac{g}{mol}

Finally, the molar mass of the sample of gas is 9.17 \frac{g}{mol}.

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How can you distinguish between crystalline allotropic modifications of Sulphur from those of amorphous allotrops?​
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The crystalline allotropes of sulfur are very strong and have a high melting and boiling point while the amorphous allotropes of sulfur are brittle and breaks easily.

<h3>What is a crystalline substance?</h3>

A crystalline substance is one that has a definite arrangement of the atoms in the substance. An amorphous substance lacks this definite arrangement. We can see this arrangement when we conduct an X-ray crystallography of the sulfur.

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4 0
2 years ago
An alloy is a mixture that has
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an alloy is a mixture of elements that has metallic properties

Explanation:

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3 years ago
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1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is
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3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

\frac{6}{80}=\frac{V_{2} }{40}

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So, 3 L will be the final volume for the gas as per Charle's law.

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3 years ago
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