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3 years ago

ICl has a higher boiling point than Br2. What is the best explanation for this?

Chemistry

2 answers:

Burka [1]3 years ago

8 0

Answer:

b. ICl experiences dipole-dipole interactions

Explanation:

The molecule with the stronger intermolecular forces will have the higher boiling point.

In Br₂, the Br-Br bond has a dipole moment of zero, because the two atoms are identical.

In ICl, the I-Cl bond, has two different atoms. One must be more electronegative than the other, so there will be a non-zero bond dipole.

ICl will have the higher boiling point.

a is wrong. Br₂ is nonpolar, so it has no dipole-dipole interactions.

c is wrong. Br₂ cannot form hydrogen bonds, because there is no hydrogen.

d is wrong. ICl has dipole-dipole interactions.

77julia77 [94]3 years ago

6 0

Answer:

B. ICl experiences dipole-dipole interactions.

Explanation:

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Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

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If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

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