CHEMISTRY COMMUNITY

Jessica Huang 1M

The C given in the problem is in units of kJ/C. You want your q to be in units of Joules or kJ. If you used the q=mC(delta T) with your given C, your q would be in units of (grams)(kJ). In the problem, you were given the heat capacity, not the specific heat capacity. Therefore, you don't need mass t...

Jessica Huang 1M

Addition: you start off with two molecules for reactants, you end up with one molecule in products.

Substitution: you start off with two molecules for reactants, you end up with two molecules in products.

Jessica Huang 1M

Ea= deltaH + RT is just an approximation. To find the actual Ea, (deltaG of activation)= (deltaH of activation) -T(deltaS of activation). But, in practice the deltaG equation is difficult to use because it's difficult to know what the activated complex is. Therefore, in reality, one would use Ea= de...

Jessica Huang 1M

Jessica Huang 1M

second order process means that rate=k[NO2]^2 must find [NO2]: 0.42 g NO2*(mol NO2/46 g)/0.15 L = 0.06087 mol NO2/L rate= k[NO2]^2 =(0.54 L/mol/s)* (0.06087 mol NO2/L)^2 = 0.0020 mol/l/s = initial rate For second part of the question, repeat above process with new initial amount of NO2 and then comp...

Jessica Huang 1M

"In the process, the temperature of the gas increased to 28.1 degrees C." During the process, volume decreased from 3.00 L to 0.500 L. Volume did not remain constant, so you cannot use Cv. The solution works because entropy is a state function. You can calculate the change in entropy when ...

Jessica Huang 1M

I don't think delta S is dependent on whether or not the process is reversible or irreversible. In both cases volume increased for the gas, which means disorder of the gas molecules increases (delta S increases).

Jessica Huang 1M

Jessica Huang 1M

It's a maximum theoretical yield. In real life, things tend to go toward entropy. When doing experiments, there's always a small percentage of error.

Jessica Huang 1M

Well, the water took 0.5 hours and 3135 J. The problem is saying that the ice took 10 hours to melt. There are 20 half hours in the period of 10 hours. In 1 hour, there are 2 half hours. (10 hours)*(2 half hours/1 hour)= (10 hours/ 0.5 hours) = 20 half hours. Water took 3135 J in half an hour. Ice w...

Jessica Huang 1M

Standard state of a liquid is a liquid. Standard state of a solid is a solid. Standard state of an element is whatever state it is found in at 1 atm and at 25 degrees C. For example, the standard state of bromine is liquid because at 1 atm and 25 degrees C, bromine is found as a liquid.

Jessica Huang 1M

When doing titrations, H+ and OH- ions would combine to form H2O. The bond formation of H2O gives off heat, heat under a constant pressure (enthalpy).

Jessica Huang 1M

Na+ and Cl- are not strong enough to change H2O on a molecular level. Meaning that OH- or H3O+ are not produced when NaCl is added. Therefore the pH does not change.

Jessica Huang 1M

Jessica Huang 1M

I don't think so. To solve that problem, you would need a graphing calculator and we are only allowed scientific calculators.

Jessica Huang 1M

Jessica Huang 1M

Jessica Huang 1M

h/p and hp^-1 are the same thing. p^-1 puts p in the denominator.

Jessica Huang 1M

It would be best to draw the most stable structure (based on formal charge).

Jessica Huang 1M

If you're not part of the solution, then you're part of the precipitate.

Jessica Huang 1M

based on number of regions of electron density

Jessica Huang 1M

Are you made of Fluorine, Iodine, and Neon? 'Cause you are F-I-Ne

Jessica Huang 1M

Before between 2 and 1.5, one has to look specifically at the heteronuclear atoms. Case by case basis.

Jessica Huang 1M

You can use either one. I personally use the E=-hR/n^2 because I understand what I am looking for. That equation would tell me the energy of the specific energy level and I can follow that. I don't like using the Rydberg equation because I don't understand how I can get from energy levels to wavelen...

Jessica Huang 1M

Converting kJ/mol to joules: 348 kJ/mol * (1000 J/ 1 kJ) * (1 mol of C-C bonds/6.022x10^23 C-C bonds) = 5.78x10^-19 J Plug 5.78x10^-19 J for E in the equation E=hc/(lamda). Solve for lamda. You would get 3.44x10^-7 m or 344 nm. Visible light wavelength is within the range of 700nm and 400 nm. Theref...

Jessica Huang 1M

CHEMISTRY COMMUNITY

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