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Orlov [11]
4 years ago
12

Water freezes at 0∘C and CO freezes at −205∘C. Which type of intermolecular force accounts for this difference in freezing point

between the two compounds?
Chemistry
1 answer:
Assoli18 [71]4 years ago
6 0

Explanation:

The dipoles in CO are in opposite directions so they cancel each other out, although CO₂ has polar bonds, it is a nonpolar molecule. Therefore, the only intermolecular forces are London dispersion forces. Water (H2O) has hydrogen bond present which is a polar bond which has a high intermolecular force.

Water which has high intermolecular force will require more energy that is a higher temperature to overcome these attractions and are pulled together tightly to form a solid at higher temperatures, so their freezing point is higher.

As the temperature of a liquid decreases, the average kinetic energy of the molecules decreases and they move more slowly.

CO with lower intermolecular forces will not solidify until the temperature is lowered further.

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How much energy is required to change the temperature of 35.0 g of Lead
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Answer:

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Explanation:

Heat Transfer Equation of pure condensed phase substance => Q = mcΔT

Mixed phase (s ⇄ l melting/freezing, or l ⇄ g boiling/condensation) heat transfer equation => Q = m∙ΔHₓ; ΔHₓ = phase transition constant

Since this is a pure condensed phase (or, single phase) form of lead (Pb°(s)) and not melting/freezing or boiling/condensation, one should use

Q = m·c·ΔT

m = mass of lead = 35.0g

c = specific heat of lead = 0.16J/g°C

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Calcium atoms have 20 protons, but tend to form ions with a 2+ charge. How
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3 years ago
Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/
Ivenika [448]

Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

                                  x = 0.0727 mol of Hydrogen

For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

                                  16 g of Oxygen ------------- 1 mol of O

                                  0.3673 g ---------------------   x

                                   x = (0.3673 x 1)/ 16

                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

                                        C₃H₃O

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