Answer:
The change in the internal energy of the system -878 J
Explanation:
Given;
energy lost by the system due to heat, Q = -1189 J (negative because energy was lost by the system)
Work done on the system, W = -311 J (negative because work was done on the system)
change in internal energy of the system, Δ U = ?
First law of thermodynamics states that the change in internal energy of a system (ΔU) equals the net heat transfer into the system (Q) minus the net work done by the system (W).
ΔU = Q - W
ΔU = -1189 - (-311)
ΔU = -1189 + 311
ΔU = -878 J
Therefore, the change in the internal energy of the system -878 J
c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.
The structure of leucine is CH3CH(<u>CH3</u>)CH2CH(NH2)COOH.
The structure of isoleucine is CH3CH2CH(<u>CH3</u>)CH(NH2)COOH.
In leucine, the CH3 group is <em>two carbons away</em> <em>from</em> the α carbon; in isoleucine, the CH3 group is on the carbon <em>next to</em> the α carbon.
Thus, <em>isoleucine</em> has the closer branched carbon.
“One is charged, the other is not” is i<em>ncorrect</em>. Both compounds are uncharged.
“One has more H-bond acceptors than the other” is <em>incorrect</em>. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.
“They have different numbers of carbon atoms” is <em>incorrec</em>t. They each contain six carbon atoms.
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:
The ICE table is then shown as:
Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
![K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}](https://tex.z-dn.net/?f=K_a%20%20%3D%20%5Cfrac%7B%5BC_3H_6ClCO%5E-_2%5D%5BH%5E%2B%5D%7D%7B%5BC_3H_6ClCO_2H%5D%7D)
where ;
Since the value for 
is infinitesimally small; then 1.8 - x ≅ 1.8
Then;
Dissociated form of 4-chlorobutanoic acid = 
Percentage dissociated = 
Percentage dissociated = 
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
Hi,
The statement is true, as the volume of a sample depends on its size.
I hope this helps. If I was not clear enough or if you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistakes.
