Use the drop-down menus to name the following structures.
2 answers:
Structure 1: 2-Butyne
Parent chain contains 4 carbons and a triple bond. So, the parent name is Butyne. Position of triple bond is written 2, as triple bond starts at carbon 2.
Structure 2: <em>cis</em>-2-Pentene
Parent chain contains 5 carbons and a double bond. So, the parent name is Pentene. Position of double bond is written 2, as double bond starts at carbon 2. As both substituents are on same side of double bond, so cis depicts the stereo.
Structure 3: <em>trans</em>-3-Decene
Parent chain contains 10 carbons and a double bond. So, the parent name is Decene. Position of double bond is written 3, as double bond starts at carbon 3. As both substituents are on opposite side of double bond, so trans depicts the stereo.
Structure 3: 1-Hexyne
Parent chain contains 6 carbons and a triple bond. So, the parent name is Hexyne. Position of triple bond is written 1, as triple bond starts at carbon 1.
Parent chain contains 4 carbons and a triple bond. So, the parent name is Butyne. Position of triple bond is written 2, as triple bond starts at carbon 2.
Structure 2: <em>cis</em>-2-Pentene
Parent chain contains 5 carbons and a double bond. So, the parent name is Pentene. Position of double bond is written 2, as double bond starts at carbon 2. As both substituents are on same side of double bond, so cis depicts the stereo.
Structure 3: <em>trans</em>-3-Decene
Parent chain contains 10 carbons and a double bond. So, the parent name is Decene. Position of double bond is written 3, as double bond starts at carbon 3. As both substituents are on opposite side of double bond, so trans depicts the stereo.
Structure 3: 1-Hexyne
Parent chain contains 6 carbons and a triple bond. So, the parent name is Hexyne. Position of triple bond is written 1, as triple bond starts at carbon 1.
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Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)
That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.