An example of erosion is the Grand Canyon, which was worn away over time by the Colorado river.
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.
A. NaCl(s) and O2(g)
B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)
C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3
D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)
E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl
F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2
G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2
H. Percent yield = 10/45.1 • 100% = 22.2% yield
Answer:
melting, freezing, sublimation, deposition, condensation, and vaporization. These changes
Explanation:
hope it helps...i dont know if its right but i hope it helps