Answer: Derived lipids:
Hydrolysis product of simple and compound lipids is called derived lipids. They include fatty acid, glycerol, sphingosine and steroid derivatives. Steroid derivatives are phenanthrene structures that are quite different from lipids made up of fatty acids.
Explanation:
Answer:
280 g Al₂O₃
Explanation:
To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).
Molar Mass (Al₂O₃): 101.96 g/mol
2.7 moles Al₂O₃ 101.96 g
------------------------ x ------------------- = 275 g Al₂O₃ = 280 g Al₂O₃
1 mole
The question is incomplete, here is the complete question:
The rate of certain reaction is given by the following rate law:
![rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=rate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
At a certain concentration of ![H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=H_2%20and%20%5Btex%5DI_2%2C%20the%20initial%20rate%20of%20reaction%20is%200.120%20M%2Fs.%20What%20would%20the%20initial%20rate%20of%20the%20reaction%20be%20if%20the%20concentration%20of%20%5Btex%5DH_2%20were%20halved.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EAnswer%20%3A%20The%20initial%20rate%20of%20the%20reaction%20will%20be%2C%200.03%20M%2Fs%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3EExplanation%20%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3ERate%20law%20expression%20for%20the%20reaction%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%5Btex%5Drate%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
As we are given that:
Initial rate = 0.120 M/s
Expression for rate law for first observation:
![0.120=k[H_2]^2[NH_3]](https://tex.z-dn.net/?f=0.120%3Dk%5BH_2%5D%5E2%5BNH_3%5D)
....(1)
Expression for rate law for second observation:
![R=k(\frac{[H_2]}{2})^2[NH_3]](https://tex.z-dn.net/?f=R%3Dk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D)
....(2)
Dividing 2 by 1, we get:
![\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%7D%7B0.120%7D%3D%5Cfrac%7Bk%28%5Cfrac%7B%5BH_2%5D%7D%7B2%7D%29%5E2%5BNH_3%5D%7D%7Bk%5BH_2%5D%5E2%5BNH_3%5D%7D)
Therefore, the initial rate of the reaction will be, 0.03 M/s
I dont know what subject is this
Answer:
F i think i pretty sure F!!!!!!!!
