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jeyben [28]
3 years ago
7

A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. T

he heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.
Chemistry
1 answer:
zubka84 [21]3 years ago
6 0

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: P_{1} = 2.7 atm,   V_{1} = 3.1 L,     T_{1} = 300 K

P_{2} = ?,      V_{2} = 9.4 L,       T_{2} = 610 K

Formula used to calculate the final temperature is as follows.

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

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm

Now, moles present upon heating the cylinder are as follows.

P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.

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The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
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Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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