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Hibiscus I
Hi yes
Grease is nonpolar. It is made up of lengthy hydrocarbon chains attached to glycerol (triacylglycerols). Because of this property, nonpolar solvents, such as tetrachloroethylene, can get grease out of clothes. This solvent acts by dissolving the grease, thereby removing it from clothes without leaving any residue.
35 b.
37 h
39 i
41 d
hope that helps
Answer:
the mesopelagic, dysphotic, or twilight zone
Explanation:
Marine zones are the divisions of the ocean. The ocean is divided into two basic parts; the pelagic or open ocean, and the benthic or sea floor.
The pelagic zone is further divided into five broad zones according to how far down sunlight penetrates and they are:
1) the epipelagic, euphotic, or sunlit zone: the top layer of the ocean where enough sunlight penetrates for plants to carry on photosynthesis.
2) the mesopelagic, dysphotic, or twilight zone: a dim zone where some light penetrates, but not enough for plants to grow.
3) the bathypelagic, aphotic, or midnight zone: the deep ocean layer where no light penetrates.
4) the abyssal zone: the pitch-black bottom layer of the ocean; the water here is almost freezing and its pressure is immense.
5) the hadal zone: the waters found in the ocean's deepest trenches.
Answer:
1) 950 mL
2) 625 mmHg
3) 426 mL
Explanation:
1) This is the relationship between pressure and volume. This relationship looks like this:
P1*V1 = P2*V2
This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.
1 L * 1 atm = 1.053 atm * X
X = 0.95 L or 950 mL
2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.
750 mmHg * 0.25 L = 0.3 L * X
X = 625 mmHg
3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.
V1 / T1 = V2 / T2
0.4 L / 303 K = X / 323 K
X = 0.426 L pr 426 mL
These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).
Good luck with you studies! :)
