Hello Hope, In this problem you are titrating a weak acid (acetic acid, CH3COOH) with a strong base (sodium hydroxide, NaOH). The acid ionization constant (Ka) for acetic acid is given as 1.76x105. The -log[Ka] = pKa = 4.75. Before the titration...

Hello Hope, In this problem you are titrating a weak acid (acetic acid, CH3COOH) with a strong base (sodium hydroxide, NaOH). The acid ionization constant (Ka) for acetic acid is given as 1.76x105. The -log[Ka] = pKa = 4.75. Before the titration...

Hello Vivian, To calculate the activation energy of a reaction, we use the Arrhenius equation. You may want to look it up to see how and why it works. In the problem you posted, there are two temperatures and two rate constants. After some rearranging and substitution...

Hello Vivian, The integrated rate equation for a second order reaction is 1/[A]t-1/[A]o=kt, where [A]t = concentration after time (t), [A]o = initial concentration, k = rate constant, & t = time (s) What you have here is a plug and chug. Let's...

Chemistry, I need some help (answer)

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...

At 1 atm, how much energy is required to heat 77.0 g of H2O(s) at –12.0 °C to H2O(g) at 121.0 °C? (answer)

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...