Answer:
The hot water was better for removing the oil.
Explanation:
You can see that because the mass went down more with the hot water. So, that means that more oil was taken out of the feather with hot water.
Nuclear fusion and nuclear fission are two different types of energy-releasing reactions that occur in the nuclei of an atom.
Here are the major differences between the two:
1. To differentiate the two, fission is the splitting of an atom into two or more smaller atoms while fusion is the conjoining or fusion of two or smaller atoms into larger one.
2. Fission does not normally occur in nature while fusion occurs mostly in heavenly bodies such as the stars.
3.Fission produces highly radioactive particles that can be hazardous to both the living things and its habitat or environment while fusion is "clean energy" and "environmental friendly" meaning there are fewer radioactive particles are produced. But if a fission "trigger" is being used, there will be radioactive particles produced.
Among the two nuclear changes, fission is widely used because this reaction produces heat in nuclear reactor. This heat is used to generate steam which operates the turbines to eventually produce electricity.
Answer:
It is false :l
Explanation:
Pahoehoe is a smooth and continuous lava crust. Pahoehoe forms when the effusion rate is low and consequently the velocity of lava flow is slow. Pahoehoe lava flow is usually at least 10 times slower than typical aa lava flow.
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Review and Study Material Before Going to
Class.
Seek Understanding.
Take Good Notes.
Practice Daily.
Take Advantage of Lab Time.
Use Flashcards.
Use Study Groups.
Break Large Tasks Into Smaller Ones.
<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28MgCl_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Mg%28OH%29_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCl%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-641.8%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%281%5Ctimes%20%28-924.5%29%29%2B%282%5Ctimes%20%28-92.30%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-16.3kJ)
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
