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Sangita_Sub_3H

For this question, the solution in the course reader says q = m*C*delta T + 1/2*m*(enthalpy of vaporization). Why is the enthalpy of vaporization multiplied by 1/2?

Sangita_Sub_3H

What was the charge when NaCl was arrested?

A salt!

Sangita_Sub_3H

In addition, the more positive the value is for the standard reduction potential, the more likely the substance is going to be reduced.

Sangita_Sub_3H

When I read the Halides section of the book again, it did not say anything about the halides getting priority over the triple bond. So, that's why I think the answer is the way it is in the textbook. Hope this helps!

Sangita_Sub_3H

I think it's because it is easier to use whole numbers instead of fractions, since half of 2x would be x. However, I think both ways should yield the same result.

Sangita_Sub_3H

I think to write the overall rate law, the equation would be 2ClO(g)+ O2(g) -->2ClO2(g), in which the O2 from step one and the 2O2 from step two would be cancelled and become 1 O2(g). Therefore, the rate law would be rate = k[ClO2]2[O2].

Sangita_Sub_3H

However, neo can only be used for pentane and up while tert can apply to butane.

Sangita_Sub_3H

Ka = sqrt (K) because Ka written out is Ka = ([H][F])/[HF] and K = ([H]^2[F]^2)/[HF]^2 since the equation is 2HF(aq) -->2H(plus) + 2F(minus).

Sangita_Sub_3H

Hi! So for part a) of this problem, the equation is AgBr --> Ag(plus) + Br (minus). So for the half reactions, you look at how the oxidation state of each reactant changes. In the case of Ag, the oxidation state changes from 0 to +1, which means that it is being oxidized since it is losing one elect...

Sangita_Sub_3H

Hi! Okay so for part a of this problem you have the equation AgBr(s) -->Ag(+) + Br-(aq). So in this case you have the two half reactions of Ag(s) --> Ag(+) +e- and Br + e- -->Br-(aq). You can see that Ag is being oxidized and Br is being reduced based on the reaction either losing or gaining electro...

Sangita_Sub_3H

In addition, when bonds are formed, the bond enthalpies are negative, and when the bonds are broken, the bond enthalpies are positive. So, you can then just add up the new bonds formed and broken.

Sangita_Sub_3H

Hi! So, since you know that 0 energy is lost to the surroundings, you then set up the primary equation q(copper)+ q(water) = 0, where q = m*C*delta T. However, since you are solving for temperature, the equation for copper would be q = (20.0g)*(0.38 J/(g*degrees C))*(T - 100 degrees C) and the equat...

Sangita_Sub_3H

Hi! When the Kc increases along with the temperature, the reaction is endothermic because Kc and temperature have an inverse relationship for exothermic reactions, whereas Kc and temperature have a directly proportional relationship for endothermic reactions. Hope this helps!

Sangita_Sub_3H

I think when in doubt, it is always best to keep 3 sig figs, even for ph :)

Sangita_Sub_3H

Hi! So, in order to figure out the overall reaction for polyprotic substances, it would probably be easier to use two steps in order to balance the overall reaction equation and to figure out the K overall. However, two steps are not necessary; if you know how many H+ ions the substance accepts or d...

Sangita_Sub_3H

Hello! So, you are given the initial pressures for each of the elements in the reaction. Using these initial values, you need to set up an ICE chart to find the x values within the equilibrium pressures of each element. In order to find x, you set the K value (given) equal to the expression of K wit...

Sangita_Sub_3H

Hello!

So the picture below should demonstrate how to solve the problem-- I had to use an ICE box in order to solve the problem. Hope this helps!

Sangita_Sub_3H

Hi! Although Q and K are calculated the same way, the same numbers are not used. For Q, the numbers that are used for the calculation are not at equilibrium; rather, they are measured values at a specific point in time. However, when measuring K, the values used are at equilibrium, which is why Q an...

Sangita_Sub_3H

Another thing that can help you is looking at the formal charge of each of the elements in the compound. Since you want each atom to have a formal charge of zero, you can rearrange the bonds accordingly.

Sangita_Sub_3H

Potassium Phosphide is written as K3P. 1. First, you see that there are 3 K atoms and I P atom. 2. For K, there is only 1 valence electron, so multiplying one by three, you get 3 valence electrons for K. 3. For P, there are 5 valence electrons. Adding the P valence electrons to the K, you get 8 vale...

Sangita_Sub_3H

It is not possible to convert Hz to m^-1. However, Hz is equivalent to s^-1.

Hope this helped!

Sangita_Sub_3H

But lets say you are removing the first two electrons from Magnesium since it has two electrons in the outer shell. After you remove the first electron, there is only one left; is it easier to remove that second electron because Magnesium wants to be in a balanced state? Or is it still more difficul...

Sangita_Sub_3H

Dear Joslyn, Another issue that I found with your calculation was with the hydrogen. After you divided the 7.88g of hydrogen given by 1.008 (the molar mass of the hydrogen element), you got the number 7.30 mol, whereas I re-did the calculation and got 7.82 mol. Then, after dividing by the lowest mol...

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