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astra-53 [7]
3 years ago
10

What do navigators need to consider when plotting a course?

Chemistry
1 answer:
Vaselesa [24]3 years ago
8 0
1.) scale of the chart
2.) Notes of the chart
3.) chart symbols
4.) chart corrections
5.) GPS positions
6.) radar fixes

7.) Visual fixtures and position circle and position line
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Some people are so good at their professions that other companies will try to recruit them?
Stella [2.4K]
Because they are good at their professions
7 0
3 years ago
You are a chemist working in your laboratory. In your storage closet you have a cabinet that holds only samples of pure elements
Vladimir79 [104]

Answer:

Mercury

Explanation:

While it is true that most metals are solid at room temperature, mercury is liquid at room temperature hence mercury is often designated as the 'liquid metal'.

Thus, if i find a bottle on the shelf that has no solid in it, only liquid and i know that only pure metals are stored in that area of the laboratory, then i will quickly relabel it as mercury.

7 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

6 0
3 years ago
HELPPP TODAY IS THE LAST DAY OF SCHOOL
Alecsey [184]

Answer:

orbiting closer to the earth's surface.....im pretty sure abt it

4 0
3 years ago
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • 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
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

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

0.015733 g Au ≈ 0.0157 g Au

3 0
3 years ago
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