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Naya [18.7K]
2 years ago
14

Which of the following is not an example of potential energy? Group of answer choices electrical energy, chemical energy, gravit

ational energy, elastic energy.
PLS HELP FAST
Chemistry
1 answer:
Sholpan [36]2 years ago
6 0

Answer: The one listed below that's NOT an example of potential energy is mechanical energy. Mechanical energy is categorized as a kinetic energy with light, sound, and thermal/heat energy.

HOPE THIS HELPS

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Paraffin oil is used for determination of boiling point and melting point for the following reasons: It has a very high boiling point and so it can be used to maintain high temperatures in the boiling and melting point apparatus without loss of the substance.

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What is specific gravity in minerals?
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Specific gravity is the "heaviness" of a mineral. It is defined as a number that expresses the ratio between the weight of a mineral and the weight of an equal volume of water.

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Brainiest if answered in the next 5m
bearhunter [10]
The equation is already balanced

2AgNO3 + MgCl2 => 2AgCl + Mg(NO3)2
7 0
3 years ago
6.) 50.0 mol H2O<br> ? molecules
8_murik_8 [283]
<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<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>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

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

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} ) = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

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

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0
2 years ago
What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in
Viktor [21]

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

3 0
2 years ago
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