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anonymous
 one year ago
If 45g of LiF dissolved in 1.8kg of water, what would be the expected change in boiling point? The boiling point constant for water (Kb) is 0.51 degrees C/m.
 0.49 degrees C
 0.98 degrees C
 1.9 degrees C
 3.5 degrees C
Im not sure which one it is I got both 0.49 and 0.98 Please Help. I will medal.
anonymous
 one year ago
If 45g of LiF dissolved in 1.8kg of water, what would be the expected change in boiling point? The boiling point constant for water (Kb) is 0.51 degrees C/m.  0.49 degrees C  0.98 degrees C  1.9 degrees C  3.5 degrees C Im not sure which one it is I got both 0.49 and 0.98 Please Help. I will medal.

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aaronq
 one year ago
Best ResponseYou've already chosen the best response.2What calculations did you perform?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I ended up going with the first answer 0.49 based on my calculations

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2did you have \(\large \Delta T=i*K_b*m=2*(0.51~^oC/m)*\dfrac{(\frac{45~g}{25.9~g/mol})}{1.8~kg}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No and when I solve yours I get 25 which isn't a choice

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2you're solving it incorrectly, because i get 0.984555984555984583356

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know who's right

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2I'm right lol you got 0.49 because you didn't take into account that i=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay thank you so much

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2no problem, and sure, shoot.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 65 gram of some unknown metal at 100.0 deg C is added to 100.8grams at 22 degrees Celsius. The temperature of the water rises 27.0 degrees C. Of the specific heat capacity of liquid water is 4.18 J/degC*g, what is the specific heat of the metal?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The choices are 2.25 J/(C*g) 1.75 J/(C*g) 0.444 J/(C*g) 0.324 J/(C*g)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know Where to start @arronq and how to get the answer

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2So this is a little complicated but bare with me. First you use the mass of the water, the specific heat capacity and the change of temperature the water had to find the heat lost by the metal. Now that you have the heat lost by the metal, you can use the mass of the metal and the temperature change of the metal (along with the heat you found in the first part) to solve for the specific heat capacity. The equation is \(\sf q=m*C_p*\Delta T\) q= heat m=mass Cp=specific heat capacity \(\Delta T\)=change in Temp

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Based on what your saying and what I do know I'm getting 0.444 with my calculations. Is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't I don't have a camera and my computers to old to type them in

aaronq
 one year ago
Best ResponseYou've already chosen the best response.2so how am i supposed to see if you're doing it right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I am I just need to double check the final response with you and if I get it wrong I'll try it a different way until I'm doing it right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay thank you so much
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