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Elena-2011 [213]
3 years ago
11

You're at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of

a lion. Startled, you accidentally release the balloon. It flies away. By the time it reaches the clouds, the atmospheric pressure that high is only 0.300 atmospheres. What would be the volume of the balloon up there?
Chemistry
1 answer:
fomenos3 years ago
8 0

Answer: The volume of the balloon up there is 6.192 L.

Explanation:

Given: V_{1} = 1.80 L,    P_{1} = 785 mm Hg (mm Hg = 0.00131579) = 1.032 atm

P_{2} = 0.300 atm,       V_{2} = ?

Formula used to calculate volume is as follows.

P_{1}V_{1} = P_{2}V_{2}

Substitute the value into above formula as follows.

P_{1}V_{1} = P_{2}V_{2}\\1.032 atm \times 1.80 L = 0.300 atm \times V_{2}\\V_{2} = 6.192 L

Thus, we can conclude that the volume of the balloon up there is 6.192 L.

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Chromium has an atomic mass of 51.9961 u and consists of four isotopes, 50 Cr , 52 Cr , 53 Cr , and 54 Cr . The 52 Cr isotope ha
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Answer : The atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

Explanation :

We know that:

Total percentage abundance of the isotope = 100 %

Percentage abundance of _{24}^{50}\text{Cr}\text{ and }_{24}^{53}\textrm{Cr}\text{ isotopes}=[100-(83.79+2.37)]=13.84\%

We are given:

Ratio of _{24}^{50}\textrm{Cr}\text{ and }_{24}^{53}\textrm{Cr} isotopes = 0.4579 : 1

Percentage abundance of _{24}^{50}\textrm{Cr} isotope = \frac{0.4579}{(0.4579+1)}\times 13.84\%=4.37\%

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = \frac{1}{(0.4579+1)}\times 13.84\%=9.49\%

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the mass of _{24}^{53}\textrm{Cr} isotope be 'x'

For _{24}^{50}\textrm{Cr} isotope:

Mass of _{24}^{50}\textrm{Cr} isotope = 49.9460 amu

Percentage abundance of _{24}^{50}\textrm{Cr} = 4.37 %

Fractional abundance of _{24}^{50}\textrm{Cr} isotope = 0.0437

For _{24}^{52}\textrm{Cr} isotope:

Mass of _{24}^{52}\textrm{Cr} isotope = 51.9405 amu

Percentage abundance of _{24}^{52}\textrm{Cr} isotope = 83.79 %

Fractional abundance of _{24}^{52}\textrm{Cr} isotope = 0.8379

or _{24}^{53}\textrm{Cr} isotope:

Mass of _{24}^{53}\textrm{Cr} isotope = x amu

Percentage abundance of _{24}^{53}\textrm{Cr} isotope = 9.49 %

Fractional abundance of _{24}^{53}\textrm{Cr} isotope = 0.0949

For _{24}^{54}\textrm{Cr} isotope:

Mass of _{24}^{54}\textrm{Cr} isotope = 53.9389 amu

Percentage abundance of _{24}^{54}\textrm{Cr} isotope = 2.37 %

Fractional abundance of _{24}^{54}\textrm{Cr} isotope = 0.0237

Average atomic mass of chromium = 51.9961 amu

Putting values in equation 1, we get:

51.9961=[(49.9460\times 0.0437)+(51.9405\times 0.8379)+(x\times 0.0949)+(53.9389\times 0.0237)]\\\\x=52.8367amu

Hence, the atomic mass of _{24}^{53}\textrm{Cr} isotope is 52.8367 amu

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