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Alex73 [517]
3 years ago
14

Define Each Piece of the Microscope

Chemistry
1 answer:
horrorfan [7]3 years ago
3 0

1) Body tube (Head): The body tube connects the eyepiece to the objective lenses. Arm: The arm connects the body tube to the base of the microscope. Coarse adjustment: Brings the specimen into general focus. Fine adjustment: Fine tunes the focus and increases the detail of the specimen.

2)Microscopes are made up of lenses for magnification, each with their own magnification powers. Depending on the type of lens, it will magnify the specimen according to its focal strength.

3)In simple magnification, light from an object passes through a biconvex lens and is bent (refracted) towards your eye. ... The eyepiece lens usually magnifies 10x, and a typical objective lens magnifies 40x. (Microscopes usually come with a set of objective lenses that can be interchanged to vary the magnification.)

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2AlF3 + 3K2O → 6KF + Al2O3
Agata [3.3K]
<h3>Answer:</h3>

7.4797 g AlF₃

<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>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

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

[RxN] 6 mol KF = 2 mol AlF₃

Molar Mass of K - 39.10 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.1 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})
  2. Multiply/Divide:                                                                                                  \displaystyle 7.47966 \ g \ AlF_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

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