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Naily [24]
4 years ago
11

The molecular weight of a gas is ________ g/mol if 3.5 g of the gas occupies 2.1 l at stp

Chemistry
1 answer:
bija089 [108]4 years ago
5 0
<span>Pre-1982 definition of STP: 37 g/mol Post-1982 definition of STP: 38 g/mol This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using. First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume. Pre-1982: 2.1 / 22.414 = 0.093691443 moles Post-1982: 2.1 / 22.71098 = 0.092466287 moles Now determine the molar mass. Simply divide the mass by the moles. So Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol Finally, round to 2 significant figures. So Pre-1982: 37 g/mol Post-1982: 38 g/mol</span>
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Answer: Option C is correct.

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Average kinetic energy is given by:

K.E._{avg}=\frac{3kT}{2}

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T = Temperature

We are given different temperatures, so to compare they all should have the same units.

a) 298K

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Looking at the temperature values, C part will have the highest average kinetic energy.

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3 years ago
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Consider the reaction: A &lt;=&gt; B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1
Katen [24]

Answer:

Keq'>1\\\Delta G'

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3

Nonetheless, we are asked for the reverse equilibrium constant that is:

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In such a way, the Gibbs free energy turns out:

\Delta G'=-RTln(Keq')\\

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

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4 years ago
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goldenfox [79]
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Answer:

Among the very basic principles that guide scientists, as well as many other scholars, are those expressed as respect for the integrity of knowledge, collegiality, honesty, objectivity, and openness.

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According to this formula:
㏑(K2/K1) = Ea/R(1/T1 - 1/T2)
when K is the rate constant
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when K is doubled so K2: K1 = 2:1 & R = 8.314 J.K^-1.mol^-1 
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㏑2 =( Ea / 8.314) (1/283 - 1/294 )
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