Put Some Numbers To It (Calculations) need calculation step by step 1) Calculate the weight of one gallon of your element (Manganese) at room temperature and atmospheric pressure. 2) Use Coloumb’s Law...

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Put Some Numbers To It (Calculations) need calculation step by step


1) Calculate the weight of one gallon of your element (Manganese) at room temperature and atmospheric pressure.


2) Use Coloumb’s Law to calculate the energy released if an electron interacts with a nucleus of your element (Manganese) at the radius of your element if they initially started out separated by a large distance.


3) Calculate the energy that would be released for a mole of such situations described in #2.


4) Calculate how long the amount of energy in #3 would be able to power a home in the USA if the average home in the USA uses 40 million BTUs of energy per year. Calculate how long it could power the average home in the world that uses 10 million BTUs of energy per year.



In The Lab (Experiment)----need more detail


1) Consider the following hypothesis: “Your element (manganese) can be separated from a random mixture ( what elements are involved in this mixture of the extract procedures) of multiple pure elements”. Design a set of procedures that could conceivably






--what is the temp ,pleas specify


isolate your element ( Manganese) from an unknown random mixture( what mixture) and subsequently identify your isolated element.



2) Describe possible outcomes of your procedures in #1. What could you conclude from each of those outcomes?


--please specify separation element and expired element


--not understanding the last answer

Answered Same DayOct 23, 2021

Answer To: Put Some Numbers To It (Calculations) need calculation step by step 1) Calculate the weight of one...

Lipika answered on Oct 24 2021
119 Votes
1. Volume of Manganese = 1 gallon = 3.785 L = 3785 cm3
Density of Manganese at room temperature and
standard pressure = 7.26 g/cm3
Volume = Molar mass/Density
So, Molar mass = Volume*Density
Here, Molar mass = 3785*7.26 g = 27,479.1 g = 27.4791 Kg
The molar mass is actually equal to the molecular weight, so, the weight of one gallon of Manganese at room temperature and atmospheric pressure is 27.4791 Kg.
2. Atomic radius of Manganese = 0.16 nm = 1.6*10-10 m
In this the electron which is negatively charged is...
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