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Lilit [14]
2 years ago
10

Where are Uranium and plutonium found? (IN NATURE)

Chemistry
2 answers:
Basile [38]2 years ago
8 0

Wanted to write more both there is a problem with the editor. Uranium is formed naturally in the crust of rocks and seawater. Plutonium does not occur in nature. It is found in the biosphere. 


evablogger [386]2 years ago
8 0
Uranium is common in the Earths crust and can be recovered in seawater. Plutonium is not found in nature but is found in naturally occurring uranium ores
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Help please!
serious [3.7K]
I think that the empirical formula is mgcl2

3 0
2 years ago
Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centip
erastovalidia [21]

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

4 0
3 years ago
What type of indications would be shown if an igneous rock cooled quickly?
Nana76 [90]

It's actually C because the rock cooled fast

6 0
3 years ago
Explain the difference between biotic and abiotic
Paladinen [302]
Biotic are living organisms and abiotic are non living things such as rocks water soil
7 0
3 years ago
A sample of 87.6 g of carbon is reacted with 136 g of
Vadim26 [7]

Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon  = 87.7g

Mass of fluorine gas  = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced  = ?

Solution:

   Equation of the reaction:

             C    +   2F₂ →   CF₄  

let us find the number of the moles the given species;

  Number of moles = \frac{mass}{molar mass}  

  C;   molar mass = 12;

            Number of moles  = \frac{87.7}{12}   = 7.31moles

 F;  molar mass  = 2(19)  = 38g/mol

             Number of moles  = \frac{136}{38}   = 3.58moles

 So;

   From the give reaction:

          1 mole of C requires 2 moles of F₂

         7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

  Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

 So;

       2 moles of F₂ will produce  mole of CF₄  

       3.58 moles of F₂ will then produce \frac{3.58}{2}  = 1.79moles of CF₄

6 0
3 years ago
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