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.
Answer: ABC&AMN are congruent by ASA comgruence
angle A=A
sideAN=AB
angle M=C
Explanation:
Answer:a line or a circle but i think circle
Explanation:
Answer:
The mass of oxygen gas dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air is 0.04936 grams.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
where,
= Henry's constant =
= partial pressure of oxygen
We have :
Pressure of the air = P
Mole fraction of oxygen in air = ![\chi_{O_2}=0.210](https://tex.z-dn.net/?f=%5Cchi_%7BO_2%7D%3D0.210%20)
![p_{O_2}=P\times \chi_{O_2}](https://tex.z-dn.net/?f=p_%7BO_2%7D%3DP%5Ctimes%20%5Cchi_%7BO_2%7D)
= Henry's constant =
Putting values in above equation, we get:
Moles of oxygen gas = n
Volume of water = V = 5 L
![Molarity = \frac{Moles}{Volume(L)}](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7BMoles%7D%7BVolume%28L%29%7D)
![0.003085 M=\frac{n}{5 L}](https://tex.z-dn.net/?f=0.003085%20M%3D%5Cfrac%7Bn%7D%7B5%20L%7D)
![n = 0.003085 M\times 5 L=0.001542 mol](https://tex.z-dn.net/?f=n%20%3D%200.003085%20M%5Ctimes%205%20L%3D0.001542%20mol)
Mass of 0.001542 moles of oxygen gas:
0.001542 mol × 32 g/mol = 0.04936 g
The mass of oxygen gas dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air is 0.04936 grams.