Answer:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
Because that solid is frozen liquid, so when it gets heated up, it will melt. Like Ice.
Hope I Helped! :)
Answer:
A catalyst is a chemical substance that alters the rate of chemical reaction not consumed by the reaction. Hence, a catalyst can be recovered chen unchanged at the ends of chemical reaction. Catalyst can be divided into two typ the basis whether it speeds up or slowdowns the rate of chemical reaction. The positive catalyst and negative catalyst.
Answer:
No
Explanation:
No, his mass remains the same no matter where he is in the universe.
But then again the moon has less gravitational pull, therefore your weight and mass will be smaller in space and on the moon than on earth
I hope this was helpful! ;)
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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 Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
![\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})](https://tex.z-dn.net/?f=%5Cdisplaystyle%20500%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%28%5Cfrac%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%7B180.18%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%7D%29%28%5Cfrac%7B2%20%5C%20mol%20%5C%20C_2H_5OH%7D%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%29)
- [DA} Multiply/Divide [Cancel out units]:
![\displaystyle 5.55001 \ mol \ C_2H_5OH](https://tex.z-dn.net/?f=%5Cdisplaystyle%205.55001%20%5C%20mol%20%5C%20C_2H_5OH)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH