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guajiro [1.7K]
3 years ago
11

A meteorologist balloon contains 250.0 L of He at 22.0 C and 740.0 mmHg. If the volume of the balloon can vary according to exte

rnal conditions what volume would it occupy at an altitude at which the temperature is -52.0 C and the pressure is .750 atm?
A. 240 L
B. 145 L
C. 760 L
D. 184790 L
Chemistry
1 answer:
NeX [460]3 years ago
5 0

Answer: Option A) 240 L

Explanation:

Given that:

Initial Volume of helium (V1) = 250.0L

Initial Temperature of helium (T1) = 22.0°C

[Convert temperature in celsius to Kelvin by adding 273

(22.0°C + 273 = 295K)]

Initial Pressure of helium (P1) = 740 mmHg

[convert pressure in mmHg to atmosphere

If 760 mmHg = 1 atm

740 mmHg = 740/760 = 0.97 atm]

Final Volume of helium (V2) = ?

Final temperature T2 = -52.0°C

[Convert -52°C to Kelvin by adding 273

-52°C + 273 = 221K]

Final pressure of helium = 0.750 atm

Since pressure, volume and temperature are given, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(0.97 x 250.0)/295= (0.750 x V2)/221

242.5/295 = 0.750V2/221

Cross multiply

242.5 x 221 = 0.75V2 x 295

53592.5 = 221.25V2

V2 = 53592.5 / 221.25

V2 = 242.2 L (Rounded to the nearest tens, V2 becomes 240 L)

Thus, the new volume of helium at the altitude is 240 liters

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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% .

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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

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negative sign is on both sides so it is canceled:

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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% .


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