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Juli2301 [7.4K]
2 years ago
15

A balloon occupies a volume of 44 L at 200K, what will the temperature of the balloon become if it is changed to 500K?​

Chemistry
1 answer:
natta225 [31]2 years ago
8 0

Answer:

110L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 44L

Initial temperature (T1) = 200K

Final temperature (T2) = 500K

Final volume (V2) =..?

The new volume of the balloon can be obtained by using Charles' law equation as shown below:

V1/T1 = V2/T2

44/200 = V2/500

Cross multiply

200 x V2 = 44 x 500

Divide both side by 200

V2 = 44 x 500/200

V2 = 110L

Therefore, the new volume of the balloon is 110L

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Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.
BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
  • Reading the Periodic Table
  • Writing Compounds
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

<u>Step 3: Use Dimensional Analysis</u>

<u />6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} ) = 511.901 g NO₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules.</em>

511.901 g NO₂ ≈ 510 g NO₂

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Answer:

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Explanation:

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