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  1. Bergmann, A., Hall, K.C., & Ross, S.M. (Eds.). (2007). Language files...

    3 Answers · Education & Reference · 03/05/2010

  2. Celsius to Kelvin: K = C + 273 Kelvin to Celsius: C = K - 273 ...

    2 Answers · Education & Reference · 21/08/2008

  3. 28.05 Grams per Mole PV = nRT K = °C + 273.15, therefore 24C = 24 + 273.15 = 297.15K 1 atm = 101325 Pa...

    1 Answers · Education & Reference · 16/12/2010

  4. I know that I used to know this years ago when I was in chemistry. C2H3O2 is acetate. That was one of the things that we had to memorize. I can't remember why or how that that is what it's called...

    2 Answers · Education & Reference · 03/06/2008

  5. You want to subtract 273 from both sides to isolate the C. True for the second part since all sides to a square are even

    2 Answers · Education & Reference · 20/08/2009

  6. The most simple answer in the quickest amount of time (coming from a mathematics major). The diameter of the circle is 8. So if you put that 8 diameter circle inside a perfect square with a 8 ft diameter, you will get an area of 64 square feet. Since the area of the circle will be...

    4 Answers · Education & Reference · 16/05/2010

  7. Area of a circle is pi*(r^2) where r is the radius. Diamater = 8 feet. So radius = diameter/2 = 8/2 = 4 pi = 3.14 (approx) so the garden's area is 3.14 * (4 * 4) = 50.24 So the answer is E.

    2 Answers · Education & Reference · 25/03/2011

  8. C = K − 273,15 K = °C + 273,15 F = K × 1,8 − 459,67 C = (°F − 32) / 1,8 K = (°F + 459,67) / 1,8 F =( °C × 1,8 ) + 32 density= m/vol there you go now think

    1 Answers · Education & Reference · 23/05/2008

  9. C = k*P then k = C/P P = 1atm C : n = P*V/RT = 1*3.10 / (0.083*283) = 0.134mol 0.134mol/1L = 0.134 M = C k = 0.134/1 = 0.134mol*L^-1*atm^-1

    1 Answers · Education & Reference · 12/12/2012

  10. Celsius to Kelvin: K = C + 273 (273.15 if you want to be precise). So 80 + 273 = 353 (or 353.15).

    3 Answers · Education & Reference · 14/01/2012

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