B. Both temperature and pressure increase
Pahoehoe is a type of
2 answers:
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Answer:
Butan-2-one
Explanation:
1. 1700 cm⁻¹
A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.
2. Triplet-quartet
A triplet-quartet pattern indicates an ethyl group.
The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.
The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.
3. 3H(s) at δ 2.13
This indicates a CH₃ group with no adjacent hydrogen atoms.
It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.
4. Identification
The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.
The compound is butan-2-one.
Answer: The volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of (molar mass = 32.0 g/mol) is as follows.
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.