Hello, this is a simple dilution problem: C1V1=C2V2, where C=concentration & V=volume Fill in the blanks: C1V1=C2V2 (0.188 M)(25.0 uL)=C2(150X10^3 uL) After converting 150.0 mL to uL, solve for C2 algebraically. C2=3...
Hello, this is a simple dilution problem: C1V1=C2V2, where C=concentration & V=volume Fill in the blanks: C1V1=C2V2 (0.188 M)(25.0 uL)=C2(150X10^3 uL) After converting 150.0 mL to uL, solve for C2 algebraically. C2=3...
Hello Amber, In a nutshell, Newton's second law states that force is equal to mass*acceleration, F=ma. The SI unit of mass is the kilogram (kg), and the SI units of acceleration are meter/second squared (m/s^2). If we multiply kg (m/s^2), we get kg m/s^2 which equals the newton,...
Helllo Brook, I expect that "logKOW" means the log of the octanol-water partition coefficient. Taking the inverse log of 3.12, we get 1318.256739. If before the extraction we have 0.10g of o-xylene in water, then post extraction we have x grams of o-xylene in octanol, and...
Hello Kaya, Using your numbers for the specific heats of solid, liquid, and gaseous water and the enthalpies of fusion and vaporization, I get 236.8 kJ of energy. Starting at -12.0 degrees C, q1=m CP dt=77.0g(2.087 J/g*C)(0+12.0 C)=1.93 kJ q2=dH of fusion=6010...