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bogdanovich [222]
2 years ago
6

Taking your foot off of the gas pedal when you see a police car which of the Newton’s law

Chemistry
1 answer:
larisa [96]2 years ago
3 0

Answer:

newton's third law of motion

Explanation:

it states about action and reaction when you see an action you take reaction like taking off your foot from the gas pedal

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DENIUS [597]

Answer:

i can't read it.. maybe upload a visible picture?

6 0
2 years ago
Which describes the path of energy that a gas powered car uses to drive along a road?
harkovskaia [24]

Answer:

mechanical energy to electrical energy to light energy

4 0
3 years ago
How many moles of ethanol are produced starting with 500.g glucose?
Monica [59]
<h3>Answer:</h3>

5.55 mol C₂H₅OH

<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 Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

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

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

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

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

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

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0
3 years ago
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
ladessa [460]

<span>Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643 1/s.
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

4 0
3 years ago
HELP!!! 30 POINTS!!!
maria [59]

Explanation:

<em><u>periodic \: law</u></em>

<h2><em><u>It's </u></em><em><u>cor</u></em><em><u>rect</u></em></h2>

5 0
3 years ago
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