A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
2 answers:
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<h3>Answer:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂
[Given] 20 mol Al₂O₃
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Al₂O₃ → 4 mol Al
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4:Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
Since our final answer already has 1 sig fig, there is no need to round.
Answer:
Explanation:
Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.
With Hess's Law in mind, let's see how A can be converted to 2C +E.
(Δ
) -----(1)
Since we have 2B, multiply the whole of II. by 2:
(2Δ
) -----(2)
This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is .
Reversing III. gives us a negative enthalpy change as such:
(-Δ
) -----(3)
This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of , which is also the equation of interest.
Adding all three together:
(
)
Thus, the first option is the correct answer.
Supplementary:
To learn more about Hess's Law, do check out: brainly.com/question/26491956