<h3>
Answer:</h3>
150 g Si
<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>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 3.2 × 10²⁴ atoms Si
[Solve] grams Si
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Si - 28.09 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>
149.266 g Si ≈ 150 g Si
Answer:
- <u><em>B. Negative association</em></u>
<u><em></em></u>
Explanation:
<em>Negative association</em> is when the independent and dependent variables move in opposite directions: if the dependent variable increases, the independent variable decreases, and if the dependent variable decreases, the independent variable increases.
As you see in the diagram, as you move to the right on the horizontal axis, meaning that the independent variable is increasing, the points lie lower, meaning that the independent variable decreases.
Therefore, if you draw a line of best fit, it will have a negative slope. A negative slope indicates <em>negative association.</em>
The graph certainly shows association as there is a clear trend; else the points would be randomly dispersded.
Hello!
The answer is C. Hibernate during the cold winter months.
Why?
Alpine marmots are known for having a long hibernation duration which starts in October (winther) and ends in April (summer) (about 7 months). During this long period, they are able to reduce their bear beats from 200 per minute to just 30 or 38 beats, and their breaths from 60 breaths/minute to 1-3 breaths/minute, guaranteeing an extreme energy saving process.
Have a nice day!
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M