You would take mass/volume to find the density. So, 12.8/2^3 is 1.6. The density of the cube is 1.6.
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: Frost weathering is a collective term for several mechanical weathering processes induced by stresses created by the freezing of water into ice.
Answer:
P₂ = 13.9 atm (3 sig. figs.)
Explanation:
The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...
P ∝ 1/V => P = k/V => k = P·V
For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂
Given:
P₁ = 1.31 atm.
V₁ = 5.51 L
P₂ = ?
V₂ = 0.520 L
V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)