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adell [148]
2 years ago
12

Wind blowing sediment against rock or land area or sand blasting is called

Chemistry
1 answer:
artcher [175]2 years ago
7 0

The correct answer is - deflation.

The process of deflation can be caused by the winds. It is an erosive process in which the main role has the wind that is carrying lot of sediment in the shape of very small particles with it.

Through this process, the winds manage to erode large areas, especially in the drier places where the vegetation is very sparsely distributed. By this type of erosion, the winds manage to make lot of hollows that can range significantly in size. The hollows made by the deflation can be anywhere from few cm deep and several meters long, up to several km long and 50-60 meters of depth.

This is the process that is responsible for the creation of most of the oasis in the largest desert in the world, Sahara, some even being lowered enough to be under the sea level.

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Draw the ring chain of isomer of propene ?​
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Explanation:

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A gaseous compound has a density of 1.90 g/L at 51.1 ∘C and 1.09 bar . Assuming ideal behavior, calculate the molar mass of the
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2 years ago
CH4 + O2 → CO2 + H2O
Scilla [17]

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Organic</u>

  • Naming carbons

<u>Stoichiometry</u>

  • Analyzing reaction rxn
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] Molar Mass of H: 1.01 g/mol

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

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

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

8 0
2 years ago
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