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Tems11 [23]
3 years ago
15

Determine the number of moles and mass of the sample of argon occupying 37.8L at STP​

Chemistry
1 answer:
Delvig [45]3 years ago
8 0

Answer:

The number of moles and mass of the sample of argon occupying 37.8L at STP​ are 1.6875 moles and 67.41 grams respectively.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then you can apply the following rule of three: if by definition of STP 22.4 L are occupied by 1 mole of Ar, 37.8 L of Ar are occupied by how many moles?

amount of moles=\frac{37.8 L* 1 mole}{22.4 L}

amount of moles= 1.6875 moles

The atomic weight of Ar is 39.948 g/mol. So the mass of the sample of argon can be calculated as:

1.6875 moles* 39.948 \frac{g}{mol} = 67.41 grams

<u><em>The number of moles and mass of the sample of argon occupying 37.8L at STP​ are 1.6875 moles and 67.41 grams respectively.</em></u>

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Transformation in chemistry is scientifically used to explain the process of changing one compound to another in a chemical reaction.

<h3>What is transformation?</h3>

The word "transformation" has a very special significance in chemistry. We know that in English, to transform would simply imply to change from one form to another. This is not quite far from its meaning in the parlance of chemistry.

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g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
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7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

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Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

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Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

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Answer:

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Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.

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Test Yourself

Identify the equation as a composition reaction, a decomposition reaction, or neither.

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