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Arada [10]
3 years ago
8

Oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5. how is charge distributed on an oxygen atom

when it bonds with a carbon atom?
Chemistry
1 answer:
vlabodo [156]3 years ago
6 0
When oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5, then the oxygen atom would have a slightly negative charge. The oxygen atom in  the carbon monoxide molecule would pull more electrons to its side since it has higher electronegativity making it slightly negative and the carbon would have a slightly positive charge as it would contain less electrons. This results to the formation of a polar molecule. A polar molecule is made when the molecule contains a slightly positive end and a slightly negative end. It would have a net dipole which is a result of the partial opposing charges in the molecule.
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Arturiano [62]
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This is the chemical formula for zinc bromate: . Calculate the mass percent of oxygen in zinc bromate. Round your answer to the
Paladinen [302]

Answer:

30%

Explanation:

<em>This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.</em>

Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂

M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)

M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol

M(Zn(BrO₃)₂) = 321.18 g/mol

Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂

There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.

6 × m(O) = 6 × 16.00 g = 96.00 g

Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂

%O = mO/mZn(BrO₃)₂ × 100%

%O = 96.00 g/321.18 g × 100% ≈ 30%

3 0
3 years ago
8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a
sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

                  Volume percentage = \frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100

                                                    = \frac{88.5}{760} \times 100

                                                    = 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

6 0
3 years ago
What is the pressure if 0.567 moles of gas occupy 4.5 liters at a temperature of 300.0 K?
kakasveta [241]
The answer is D. Use the equation PV=nRT
P=(.567mol)(.0821)(300K)/4.5L
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3 years ago
Which describes any compound that has at least one element from group 17?
Contact [7]
The answer is HALIDE.
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