<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:
A. Water particles barely move forward; they move in a circular pattern.
 
        
                    
             
        
        
        
 What  happens  when  chlorine  form  an ion  is that  it gains an electron and  has  an octet  in its  outer shell  ( answer  A)
 <u><em> Explanation</em></u>
<u><em> </em></u>Chlorine is  is in atomic  number  17  in periodic table.
The electron configuration  of chlorine  is    1S2 2S2 2P6 3S2 3P5   or[Ne]3S2 3p5  or  2.8.7.
chlorine therefore  has  7  valence electron therefore it  gain  1 electron  to form Cl- ( ion)
 Cl- has  8  electron in its outer  shell (  it  obeys  octet  rule  of eight valence in outer shell.
 
        
                    
             
        
        
        
Answer:
ΔS=0.148  KJ/K
Explanation:
Given that
Q = 100 KJ
T₁=200°C
T₁=200+273 = 437 K
T₂=5°C
T₂=5 + 273 = 278 K
Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.
So the total change in entropy given as
ΔS=  - Q/T₁ + Q/T₂
ΔS=  - 100/473 + 100/278  KJ/K
ΔS=0.148  KJ/K
 
        
             
        
        
        
The tubes 1, 5 and 6 is the answer to the question which test tubes were used to determine the optimal ph lipase activity. Lipase is an enzyme and optimum ph the maximum possible point on which enzyme become active.