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vichka [17]
2 years ago
10

Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent? Arrange the substances with polar

covalent bonds in order of increasing bond polarity:
(d) SO₂
Chemistry
1 answer:
KengaRu [80]2 years ago
3 0

Covalent bonds can be classified as nonpolar and polar covalent given the electronegativity difference between two atoms (ΔEN).

Nonpolar covalent bond electrons are shared equally between two atoms, polar covalent bond electrons are shared unequally, atoms have partial charges, ionic bond electrons are completely transferred to one atom, full charges present. Therefore, the greater the electronegativity difference, the greater the bond polarity. Let's determine the types of bonds present in the compounds and arrange the ones with polar covalent in order of increasing ΔEN. Sulfur and oxygen are both nonmetals so the substance is covalent. Sulfur has EN = 2.5 and oxygen has EN = 3.5. Since there is an electronegativity difference, the S−O bonds in the substance can be classified as polar covalent bonds.

Learn more about polar covalent bond here:

brainly.com/question/25150590

#SPJ4

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5 0
2 years ago
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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