Calculating Volume of Uranium Alloy for Critical Mass

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To calculate the volume of a uranium alloy with a density of 18.75 g/cm3 for a critical mass of 49 kg, the formula volume = mass/density is used. The smaller mass of 16 kg, when surrounded by a neutron reflector, also requires the same volume calculation approach. The discussion emphasizes that the problem's complexity is mainly due to the additional context rather than the calculations themselves. The final volume for the smaller mass can be derived from the equation 16000 g / 18.75 g/cm3. Additionally, users seek resources for learning how to draw Lewis structures, indicating a need for clearer educational materials.
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A uranium alloy has density of 18.75 g/cm3. Find the volume if occupied
by a critical mass of 49 kg of this alloy? The critical mass itself can be decreased to 16
kg if the alloy is surrounded by a layer of natural uranium ( neutron reflector). What is the volume of the smaller mass?

I am sorry but I have no idea how to do this question, I just don't get how to start , like should i use moles or possibly nuclear equations. No idea

Lastly, more importantly any good sites to teach how to draw Lewis structure
 
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You have mass, you have density, just use these number to calculate volume. That's all.
 
Borek said:
You have mass, you have density, just use these number to calculate volume. That's all.

Thanks but then what is all the decreased mass and all those

And they are asking the 'smaller' mass
 
You have two masses given - 49 kg and 16 kg. One is larger, one is smaller. There is nothing more to the question, everything else is just to make the question more interesting (or more confusing if you like).
 
Borek said:
You have two masses given - 49 kg and 16 kg. One is larger, one is smaller. There is nothing more to the question, everything else is just to make the question more interesting (or more confusing if you like).


So the answer is 16000 g /18.75g

And what will the unit for volume be
 
someone21 said:
So the answer is 16000 g /18.75g

\frac {16000 g} { 18.75 \frac {g} {mL}}

Units follow from that.
 
Borek said:
\frac {16000 g} { 18.75 \frac {g} {mL}}

Units follow from that.


Thanks for answering that but I had one more question as in my first post

Lastly, more importantly any good sites to teach how to draw Lewis structure

I know that there are many sites from google but i somehow don't find it easy to understand thereby any other good sites
 
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