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Arte-miy333 [17]
4 years ago
12

Which element is used in light bulbs as a filament

Chemistry
1 answer:
mojhsa [17]4 years ago
8 0
<span>The element that is used in light bulbs as a filament is tungsten - this is almost always the case in halogen and incandescent bulbs. Tungsten is chosen for this purpose because of the fact it can withstand temperatures of up to 4,500 degrees, as well as being incredibly flexible.</span>
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How many chlorine molecules are in 7.02 l of chlorine gas at stp?
emmasim [6.3K]
The answer is:  [B]:  1.89 * 10²³  .
____________________________________________________

PV = nRT ;  is the equation for "STP" conditions; that is, the "ideal gas equation" .
_______________________________________________
                      i.e. when Pressure, "P" = 1.00 atm; 
                                    Temperature, "T" = 273 K;
_____________________________________ 
                           R = the ideal gas constant = ((0.08206 L-atm/K-mol)
                           n = number of moles;
___________________________________
So, we plug in our known values:
______________________________________________
<span>(1.00atm) (7.02L) = ("n" mol) (0.08206 L-atm/K-mol) (273K); </span>
<span>_____________________________________________
</span>→ 7.02 L·atm = (? mol) (22.4 L·<span>atm/mol) .
</span>
     (Note that t<span>he Molar Volume of a gas at STP is a constant using Avogadro's value of </span>22.4 L / mol.  1 mol of any ideal gas at STP occupies 22.4 L. An ideal gas takes the shape of its container)..
<span>_______________________________________________________
</span> → Divide EACH side of the equation by "(22.4 L·atm/mol)" ; 
________________________________________________
→ 7.02 L·atm / = ("n" mol) (22.4 L·atm/mol) ;
_________________________________________________
→ 7.02 L·atm / (22.4 L·atm/mol) =
                     [("n"  mol) (22.4 L·atm/mol)]/(22.4 L·atm/mol);
______________________________________________________  
<span>to get:  
_________________________________________________
 </span>→ n = <span>0.313 mol ;
</span>_____________________________________________________
Note: 1 mole = 1 mol = 6.022 * 10²³  molecules.
____________________________________________
  → So, 0.313 mol Cl₂ (g) molecules * [(6.022 * 10²³  molecules) / (1 mol)] =
___________________________________________________________
  → [(0.313) * (6.022 *10²³) ] molecules of Cl₂ (g) ;
___________________________________________________________
            →  =  1.88 * 10²³ molecules of Cl₂ g ; 
___________________________________________________________
                   → which most closely corresponds with answer choice:
___________________________________________________________ 
                                                       [B]:  1.89 * 10²³  . 
___________________________________________________________  
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