Answer: They use other organisms for energy
Explanation:
Answer:
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Answer:
The answer is below
Explanation:
The separation technique is used for separating immiscible liquids.
When separating, the stopper has to be removed when draining the lower layer so as to prevent a vacuum. If vacuum is allowed, the draining rate will reduce and stop.
The liquid should be mixed by shaking the funnel and then opening the stopcock so as the vent out gases.
When near interface between the layers, you should set your eye level so that you do not drain up to the second layer.
After completely draining the first layer, the second layer should be collected in a new flask.
After mixing the solutions in a separatory funnel, the stopper should be removed and the liquid should be mixed thoroughly and the layers allowed to separate. When you get close to the interface between the layers, get eye level with the funnel and slow the draining until the first layer is collected. Switch to a new flask to collect the second layer.
<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
<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams
<u>Explanation:</u>
We are given:
44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.
To calculate volume of a substance, we use the equation:

Density of solution = 1.343 g/mL
Mass of solution = 100 g
Putting values in above equation, we get:

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:
In 77.46 mL of solution, mass of sulfuric acid present is 44 g
So, in 60 mL of solution, mass of sulfuric acid present will be = 
Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams