CHEMISTRY COMMUNITY

Kanwaldeep Rai 2H

Chem_Mod answered this question in 2013: " Ethane has more freedom due to the C-C single bond vs. ethene with a C=C double bond which is more rigid. Having more freedom means a higher heat capacity...Ethane is a more complex molecule as it contains two more hydrogens. Some of the energy when he...

Kanwaldeep Rai 2H

F is an electrophile but F- is an nucleophile.

Kanwaldeep Rai 2H

For homework question number 1.12, why do we number the diene before the methyl. I thought we learned to give the number of the methyl group priority when naming the structure?

Kanwaldeep Rai 2H

My friend and I researched this and found: The steady state method can only be used if the first step of a reaction is much slower than the second step, whereas the pre-equilibrium approximation requires the first step to be faster. I don't know if this is completely correct but this is what we found.

Kanwaldeep Rai 2H

You won't get a straight line for more than one of these because each graph represents a different variation of [B] vs time. [B], ln[B] and 1/[B] all have different plots.

Kanwaldeep Rai 2H

You can identify a second order reaction by looking at the rate is would be going. Usually, second order reaction rates are proportional to the product of two reactants, with the reactants being the same or different. Like one form of a second order reaction is 2A->C or A+B->C. The rate for the firs...

Kanwaldeep Rai 2H

The negative sign just represents what thing (water or katana) releases the heat and which absorbs the heat. Because the katana has a larger temperature, heat is going to be released from the katana to the water making q negative while the water absorbs that heat making q positive. Either of them co...

Kanwaldeep Rai 2H

Breaking a bond is always an endothermic process, but whether the reaction is endothermic or not is determined by comparing the enthalpies of the products to the enthalpies of the reactants.

Kanwaldeep Rai 2H

When bonds are formed energy is released because the molecules/atoms go from a unstable form to a stable one. They go from high energy to low energy and thus energy in terms of heat is released (exothermic). In order to break bonds, there must be an absorption of energy to overcome electrostatic att...

Kanwaldeep Rai 2H

I believe its still an isolated system because we are viewing it as an ideal situation. A thermos bottle, ideally, should not allow matter or energy exchange.

Kanwaldeep Rai 2H

I believe you can still do the problem but you would have to account for the fact that the concentration of one reactant is more than another and thus only a certain amount of H2O can be produced. I think this would become a limiting reactant problem because the mole ratio of reactant to product is ...

Kanwaldeep Rai 2H

This video explains two problems from Quiz #3. I realize that the problem may be different for other sections but its good practice. I hope this helps anyone who needed more explanation on equilibrium expressions.

Kanwaldeep Rai 2H

Oxygen, in oxalato, tends to only make two bonds from the single bonded oxygens. According to electronegativity, only the oxygens with three lone pairs attracts another atom. This is the reason that it is a bidentate and not a polydentate. I know it's confusing but that is what I was told. Also, you...

Kanwaldeep Rai 2H

For this problem, think about it terms of a system as a whole. So, when you increase the pressure or decrease the volume, the molecules are hitting the walls faster causing stress on the system- an increase in the total pressure of the system. In order to reduce the stress on the system, the reactio...

Kanwaldeep Rai 2H

When you add another gas, increasing the pressure of the gas, it does not mean that the volume has changed. You can keep the volume constant and still increase pressure. For example, if you add helium gas in a constant volume, you are only increasing the pressure.When you don't change the volume, th...

Kanwaldeep Rai 2H

My TA told my discussion section that it didn't matter which way you wrote it. Our course reader combines it into (n2P)^3 however.

Kanwaldeep Rai 2H

For BH2- the molecule shape is bent or angular because it has three regions of electron density with two bonds and one lone pair of electrons. This would result in bond angles <120 degrees. The lone pairs of electrons would repel the bonded electrons causing less of a bond angle between the actual b...

Kanwaldeep Rai 2H

Yes radical compounds usually have an odd number of valence electrons leaving one electron unpaired. An example would be CH3. However, there doesn't always need to be an odd number of valence electrons because a biradical can have an even number of valence electrons. An example of a biradical is the...

Kanwaldeep Rai 2H

Writing electron configurations depends on the energy of the shells and orbitals. The orbital with the highest energy will always be written last. That is why, for example, the "4s" orbital is written after the "3d" orbital. However, the first two periods don't have a "d&quo...

Kanwaldeep Rai 2H

The electron will come from the 4s orbital because it has more energy. More energy means more instability and thats where the valence electrons are at. It will be easier to remove an electron from a higher energized state than a lower one hence the electron for the +1 ion of Mn coming from 4s not 3d...

Kanwaldeep Rai 2H

This is correct. The Balmer Series begins with n1=2 so all the wavelengths that are listed are from n2=3,4,5,6. The next line would be n2=7. You would first find the frequency of this line. V(frequency)= R(1/n1^2-1/n2^2)= (3.29x10^15 Hz)(1/2^2-1/7^2)= 7.55x10^14 Hz Then convert this to wavelength by...

CHEMISTRY COMMUNITY

Search found 21 matches

Re: Heat Capacity

Re: Distinguishing Nucleophiles

HW Question 1.12 Numbering the structure

Re: Pre-equilibrium vs steady-state

Re: Multi-Reactant Concentration Changes

Re: Identifying second order RXNS

Re: qsys=-qsurr? Winter 2013 Midterm

Re: Endothermic or exothermic

Re: Constant P Calorimeter - Bond Formation Q???

Re: Closed versus Isolated System

Re: Measurement of Enthalpy example problem in course reader

Quiz #3 Problems 4 and 6 Explanation

Re: Bidentate or Polydentate

Re: response of equilibria to pressure as relating to moles

Re: Pressure increase by adding another gas

Re: Valence Electron Configuration

Re: Bond Angles for BH3-

Re: radical compounds

Re: Electron Configuration

Re: Electron Removal

Re: Rydberg Equation