<h3>
Answer:</h3>
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
They are called hydrocarbons
Answer:
Answer is explained below;
Explanation:
In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.
In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.
By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.
He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.
So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.
Later, the Rutherford model was replaced by the Bohr atomic model.
Answer:
4.4×10² cm³
Explanation:
From the question given above, the following data were obtained:
Diameter (d) = 68.3 mm
Height (h) = 0.120 m
Volume (V) =?
Next, we shall convert the diameter (i.e 68.3 mm) to cm.
This can be obtained as follow:
10 mm = 1 cm
Therefore
68.3 mm = 68.3 mm / 10 mm × 1 cm
68.3 mm = 6.83 cm
Therefore, the diameter 68.3 mm is equivalent 6.83 cm.
Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:
1 m = 100 cm
Therefore,
0.120 m = 0.120 m/ 1 m × 100 cm
0.120 m = 12 cm
Therefore, the height 0.120 m is equivalent 12 cm.
Next, we shall determine the radius of the cylinder. This can be obtained as follow:
Radius (r) is simply half of a diameter i.e
Radius (r) = Diameter (d) /2
r = d/2
Diameter (d) = 6.83 cm
Radius (r) =?
r = d/2
r = 6.83/2
r = 3.415 cm
Finally, we shall determine the volume of the cylinder as follow:
Radius (r) = 3.415 cm
Height (h) = 12 cm
Volume (V) =?
Pi (π) = 3.14
V = πr²h
V = 3.14 × (3.415) ² × 12
V = 440 cm³
V = 4.4×10² cm³
Therefore, the volume of the cylinder is 4.4×10² cm³
