The answer is <span>ionic aluminum fluoride (</span>AlF3). Note that boiling points of pure solvents are raised in the presence of solutes. The type of solute also affects the boiling point elevation of the solution. Ionic substances tend to raise it more than covalent ones, so sucrose is out of the picture. Next, consider the number of ions the ionic substance produces. The more ions, the greater the BPE. AlF3 dissociates into 4 ions.
Answer:
The most stable conformer would be the anti-conformer when the substituent methyl groups are farthest away from each other.
Explanation:
Isomers are chemical compounds with the same molecular formula but with different molecular structures.
Conformers are a special type of isomers that produce different structures when the substituents of a Carbon-Carbon single bond (C-C) are rotated.
In 2,3 dimethyl butane, the substituent methyl groups are located around the second and third Carbon to Carbon single bond.
To achieve a stable configuration, the methyl group substituents need to be as far apart as possible (that is, in an anti-position) to minimise repulsion.
The closer the methyl groups are to each other, the more they repel each other and the more unstable the conformer becomes.
Answer:
Br2+ 2H2O + SO2= 2HBr + H2SO4
Let's Compare the left side of the equation to the right side of the equation.
Left: Br= 2, H= 2, S= 1, O = 1+2
Right: Br=1, H= 1+2, S=1, O= 4
We can see that only S is balanced and not the other 3 elements.
I'll try to make each element balance.
For Br; I'll multiply by 2 on the left to make it equal to the right.
For H; Since the 2 for Br on the right affected also H, that H ( for HBr) Already has a 2, but then it adds with the other H2( for H2SO4) to give a total of 4 H on the right side. But then there's only 2 H on the left. so we multiply that 2 by a 2 ( which is written infront of the H2O to give a total of 4 H on the left side.
For O; Because of the 2 infront of the H2O, it affects the O in H2O..so now we have 2 O plus the 2 O ( in SO2) to give a total of 4 O which is equal to the right side.
