¿What? For you bye, i love you, no sabo ingles.
Answer:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
Answer:
They assumed they both had water.
Explanation:
Because they only could look at it through telescopes that were not advanced
Answer:
The correct answer is because the molecular structure.
Explanation:
The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.
In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.
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<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:
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In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.
On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration
The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.
The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.