Answer:
mass of Uranus: 9.31x10^22 kg
Explanation:
As I stated in the comments, the question in this problem is to calculate the mass of Uranus, given it's conditions.
Now, we have the data of the atmosphere gases, but for calculations, this data is irrelevant.
To calculate the mass of Uranus, we can use the second law of Newton which is:
F = m * a (1)
In this case, the only Force exerting here would be the weight so:
F = m * g (2)
And the force of Uranus, can also be expressed in terms of pressure and Area so:
F = P * S (3)
Now, if we replace (3) in (2) we have:
P * S = m * g
Solving for m:
m = P * S / g (4)
We have all the data, therefore, let's replace it in the above expression to calculate the mass of Uranus:
m = 1x10^8 * 8.1x10^15 / 8.7
m = 9.31x10^22 kg
This is the mass of Uranus
Because the bugs with body color A are able to blend into the habitat/trees better, which makes it harder for the birds to see
Answer:
1. False
2. False
3. True
4. False
Explanation:
1. CBr4 is more volatile than CCl4 False
The molecular weight of CBr4 is is greater than the CCl4, therefore it has less tendency to escape to the gas phase. Also, the CBr4 has greater London dispersion forces compared to CCl4 since bromine is a larger atom than chlorine.
2. CBr4 has a higher vapor pressure at the same temperature than CCl4 False
For the same reasons as above, the vapor pressure of CBr4 is smaller than the vapor pressure of CCl4
3. CBr4 has a higher boling point than CCl4 True
Again, CBr4 having a molecular weight greater than CCl4 ( 331 g/mol vs 158.2 g/mol) is heavier and less volatile with a higher boiling point than CCl4.
4. CBr4 has weaker intermolecular forces than CCl4 False
Both molecules are non-polar because the dipole moments in C-Cl and C-Br bonds cancel in the tetrahedron. The only possible molecular forces are of the London dispersion type which are temporary and greater for larger atoms.
Answer:
see explanation below
Explanation:
First, you are not providing the reaction to do the conversion. However, in picture 1, I attached a reaction that match perfectly with the description of this question, so I hope it could be.
Now picture 2, shows you the correct steps of conversion, now let me explain a little here the mechanism:
The first step in all options is the same. Those reagents works to do an oxidation of the methyl group to a carboxilic acid, and then, we transform toluene into benzoic acid. Now the second step in the first option involves Cl2 in light, and this only works to do halide alkylation in acyclic chains, so this would not work in the reaction to get the final product. In option 2, we have Cl2/FeCl3, and this will promote a friedel craft alkylation, and Chlorine will go to the para position of the ring, so we can exclude this option. However option 3 and 4 uses SOCl2, and this is often used to convert the hydroxile group of a molecule into an halide, so these two options works in the second step.
Now, for the step 3, option 4 uses grignard reagent with (CH3CH2)2CH2MgBr as reagent, but this will only form a larger chain and the CH3CH2 will not break, so, this option is discarted, and finally option 3 will be the correct one. Picture 2 shows you every step.
