There are two possible products from this elimination:
-2,3-dimethylbut-1-ene
-2,3-dimethylbut-2-ene
As the base is relatively unhindered, the reaction will form the Saytzeff product as the major product. The Saytzeff product is the most substituted alkene which is more stable due to hyperconjugation. In this reaction the Saytzeff product is 2,3-dimethylbut-2-ene.
<h3>
Answer:</h3>
388.94 g N₂
<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 Table
- Using Stoichiometry
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 4HCl (g) + 6NO (g) → 5N₂ (g) + 6H₂O (g)
[Given] 404.87 g HCl
<u>Step 2: Identify Conversions</u>
[RxN] 4 mol HCl = 5 mol N₂
Molar Mass of H - 1.01 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of N₂ - 2(14.01) = 28.02 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 388.935 g N₂
<u>Step 4: Check</u>
<em>We are given 5 sig figs. Follow sig fig rules and round.</em>
388.935 g N₂ ≈ 388.94 g N₂
Answer:
what??? ????????????????????
Properties which repeat in an order.
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
