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faltersainse [42]
4 years ago
5

Please help! Will mark Brainliest!

Chemistry
1 answer:
sergey [27]4 years ago
3 0

C) H2O has the strongest dipole interactions

Explanation:

The dipole interactions are the interactions that occur between polar molecules.

Polar molecules form when the atoms forming the molecule have different electronegativity, therefore one of the atom in the molecule attract the shared electrons more towards itself; as a result, the molecule ends up having a region which is slightly negatively charged and a region slightly positively charged.

In this problem, we are comparing 4 different molecules. We observe that:

H2: this molecule is formed by 2 atoms of hydrogen - since the two atoms are identical, electrons are equally shared between the two atoms, so this molecule is not polar

O2: this molecule is formed by 2 identical atoms of oxygen, so it is not polar, for the same reason as H2

CH4: this molecule consists of a carbon atom surrounded by 4 hydrogen atoms - since this molecule has a symmetric structure, the charge is balanced, therefore this molecule is not polar

H2O (water): the oxygen atom in the water molecule is more electronegative than the hydrogens, therefore it tends to attract more the shared electrons - as a result, the molecule is polar.

Therefore, the correct answer is H2O.

Learn more about atoms and molecules:

brainly.com/question/2757829

#LearnwithBrainly

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Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

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2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

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3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\  \\  \end{gathered}

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

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