We have that energy=specific heat * change in temperature * mass. Thus, we have the final temperature (22) minus the initial temperature (55) to equal -33 as our change in temperature. Our specific heat is in J/g*C, so we're good with that because g stands for grams and the aluminium is measured in grams. As there are 10 grams of aluminum, we have

as our final temperature
An exothermic reaction would release energy and would therefore lose heat itself, while an endothermic reaction would absorb energy and gain heat. Therefore, losing heat would be an exothermic reaction
Feel free to ask further questions!
<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:

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

<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
Answer:
The correct answer is 0.12 grams.
Explanation:
The mass of carbon monoxide or CO collected in the tube can be determined by using the ideal gas equation, that is, PV = nRT.
Based on the given question, P or the pressure of the gas is given as 1 atm, volume of the gas collected in the tube is 117 ml or 0.117 L.
The number of moles or n can be determined by using the equation, mass/molar mass.
R is the universal gas constant, whose value is 0.0821 L atmK^-1mol^-1, and temperature is 55 degree C or 328 K (55+273).
On putting the values we get:
n = PV/RT
= (1 atm*0.117 L) / (0.0821 L atmK^-1mol^-1 * 328 K)
= 0.0043447 mol
Therefore, mass of CO will be moles * molar mass of CO
= 0.0043447 mol * 28 g/mol
= 0.12 g