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coldgirl [10]
3 years ago
12

Corinne uses a model rocket in her physical science class. The rocket begins its trip traveling vertically at 22 m s . It reache

s a peak height of 24 meters and then begins its descent. Just before it hits the ground, it reaches a velocity of 22 m s . Which statement is true regarding the rocket's trip? A) The rocket travels a greater distance upward than downward. B) The rocket takes more time to travel up than to travel down. C) The rocket's velocity remains positive during the entire trip. D) The rocket's velocity on the trip down is the negative of the upward trip.
Chemistry
2 answers:
dezoksy [38]3 years ago
8 0

The answer is D have a good day

Phoenix [80]3 years ago
3 0

<u><em>Answer</em></u> : D) The rocket's velocity on the trip down is the negative of the upward trip.

<u><em>Explanation</em></u> : The rocket's velocity on the trip down is the negative of the upward trip. The trip up starts at 22 m/s and ends at 0 m/s. The trip down starts at 0 m/s and ends at -22 m/s. This assumes that up is the positive direction.

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2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

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

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

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