Answers:
1. 3-ethyl-3-methylheptane; 2. 2,2,3,3-tetramethylpentane; 3. hexa-2,4-diene.
Explanation:
<em>Structure 1
</em>
- Identify and name the longest continuous chain of carbon atoms (the main chain has 7 C; ∴ base name = heptane).
- Identify and name all the substituents [a 1C substituent (methyl) and a 2C substituent (methyl).
- Number the main chain from the end closest to a substituent.
- Identify the substituents by the number of the C atom on the main chain. Use hyphens between letters and numbers (3-methyl, 3-ethyl).
- Put the names of the substituents in alphabetical order in front of the base name with no spaces (3-ethyl-3-methylheptane)
<em>Structure 2</em>
- 5C. Base name = pentane
- Four methyl groups.
- Number from the left-hand end.
- If there is more than one substituent of the same type, identify each substituent by its locating number and use a multiplying prefix to show the number of each substituent. Use commas between numbers (2,2,3,3-tetramethyl).
- The name is 2,2,3,3-tetramethylpentane.
<em>Structure 3
</em>
- Identify and name the longest continuous chain of carbon atoms that passes through as many double bonds as possible. Drop the <em>-ne</em> ending of the alkane to get the root name <em>hexa-</em>.
- (No substituents).
- Number the main chain from the end closest to a double bond.
- If there is more than one double bond use a multiplying prefix to indicate the number of double bonds (two double bonds = diene) and use the smaller of the two numbers of the C=C atoms as the double bond locators (2,4-diene)
- Put the functional group name at the end of the root name (hexa-2,4-diene).
<em>Note</em>: The name 2,4-hexadiene is <em>acceptable</em>, but the <em>Preferred IUPAC Name</em> puts the locating numbers as close as possible in front of the groups they locate.
Branched chain alkanes
The alkanes don't contain a functional group and so the branches are numbered from the end that gives the lowest set of position numbers for the branches.
Use the above rules to see how the names of the alkanes below are built up.
The structure of 2-methylbutane is a butane molecule (C4H10) but with a methyl group (CH3) replacing a hydrogen on the second carbon atom in the chain. The structure of 3-methylpentane could be drawn as butane with an ethyl group (C2H5) replacing a hydrogen on the second carbon. Note that this is not 2-ethylbutane. The structure of 2,2-dimethylbutane is butane with two methyl groups replacing the two hydrogens on the second carbon.
<span>Well, during the day, the water, as well as the surfaces surrounding the water, are heated by various thermodynamic processes: conduction, convection, radiation, etc. This in turn warms the water molecules in the lakes, streams, rivers, and oceans, thereby transferring heat (their kinetic energy) to the water molecules, which in turn receive that energy from the surrounding surfaces, or directly via radiation/insolation from the sun. When the water molecules attain enough energy, some of them attain enough energy to escape the surface of the liquid and enter the gas phase. Hence, as water is heated, more and more water molecules attain enough kinetic energy to enter the gas phase.</span>
Answer:
B?
Explanation:
it just makes the most sense in my head
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>
<em />
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
<em>Theoretical yield:</em>
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
<em>Percent yield:</em>
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
<h3>D) 65.7%
</h3>
