Aldehydes and ketones can be halogenated at their α-position by reaction with Cl2, Br2, or I2, under acidic conditions. Using Br
2 under acidic conditions, an intermediate enol is formed which adds bromine at the α-position. The reaction stops after the addition of one bromine because the electron-withdrawing halogen decreases the basicity of the carbonyl oxygen, making the protonation less favorable. Draw curved arrows to show the movement of electrons in this step of the mechanism.
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Explanation:
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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
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<h3>Answer:</h3>
64 g O₂
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Divide/Multiply [Cancel Units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂