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Allisa [31]
3 years ago
13

In order for a roller coaster to work, why does the hill the cart climbs need to be higher than the top of the loop?

Chemistry
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

Explanation:

At the highest point of the first and tallest slope, your potential energy is at its most elevated it will at any point be on this ride. As you plunge, your potential energy diminishes until it's totally gone at the lower part of the slope." The more limited the slope the exciting ride climbs, the more prominent its active energy.

Another way of saying is that

A roller ride works utilizing gravity. the underlying slope must be sufficiently high that gravity can create sufficient force to finish the course.

Brainliest?

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1. CaCO3(s) <=> CaO(s) +CO2 (g) Delta H = +178 kJ/mol

Since Delta H is positive, it means the reaction is endothermic

a. An increase in temperature will will shift the equilibrium position to the right, thus forming more products

b. A decrease in the temperature will shift the equilibrium position to the left, thus forming more reactants

2. PCl3(g) + Cl2(g) <-> PCl5 (g) Delta H= -88 kJ/mol

From the above, Delta H is negative which implies exothermic reaction.

c. Increasing the temperature will shift the equilibrium position to the left, thus forming more reactants.

d. Decreasing the temperature will shift the equilibrium position to the right, thus forming more products

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How many grams of CO2 are produced from 6.7 L of O2 gas at STP?
Tasya [4]
<h3>Answer:</h3>

13 g CO₂

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Using Dimensional Analysis

<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>
<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] 6.7 L O₂

[Solve] g O₂

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

[STP] 22.4 L = 1 mol

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

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

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

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 13.1637 \ g \ O_2

<u>Step 4: Check</u>

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

13.1637 g CO₂ ≈ 13 g CO₂

5 0
3 years ago
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