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Makovka662 [10]
3 years ago
8

The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi

de the blimp
Chemistry
1 answer:
anygoal [31]3 years ago
8 0

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

  • P = 1 atm
  • V = 5.7x10⁶ L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 25 °C ⇒ (25+273.16) = 298.16 K

And we <u>solve for n</u>:

  • 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • n = 4.29x10⁻⁶ mol

Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:

  • 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

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2. 90 mg

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O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

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1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

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