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poizon [28]
3 years ago
6

What is the effect of the room temperature on how well s student does on a test

Chemistry
1 answer:
yarga [219]3 years ago
4 0
Temperature has been shown to have a tremendous impact on student performance, and it all begins with altering our mood. Temperature’s effect on mood decreases memory and cognitive ability which is what affects learning ability. Data collected from a study done by the University of Scranton reveals that students perform best in controlled temperatures with test scores averaging in the 90th percentile. When temperatures were too hot scores averaged in the low 70s, and at the mid-70s when they were too cold.
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Classify each of the following molecules as purely covalent, non-polar covalent, polar covalent,
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A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

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