The specific heat of water is 4.18 J/(g * °C) How much heat, in kilojoules, must be added to 250 g of water to increase the temperature of the water by 5.0°C?

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Final answer:

The solution with the highest number of particles in solution (CaCl2 in this case), experiences the highest osmotic pressure, lowest vapor pressure and highest boiling point due to the principles of colligative properties.

Explanation:

The question pertains to the colligative properties of solutions, which would be governed by the number of particles in the solution. The solutions are 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3, and 0.04 m CaCl2. For (i) Highest osmotic pressure, the solution with the highest ion count would yield the highest osmotic pressure. CaCl2 dissociates into three ions (Ca²+, and 2 Cl¯), therefore, it would exhibit the highest osmotic pressure. For (ii) Lowest vapor pressure, this would coincide with the solution with the highest osmotic pressure, again making it CaCl2, due to the greatest decrease in vapor pressure. For (iii) the highest boiling point, this too would be CaCl2 for the reasons stated above. The presence of more particles in a solution interferes more with the evaporation process, requiring more energy (higher temperature) to achieve boiling.

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Answer:

555 ft per min

Explanation:

For the problem above, we can use the IAP chart and the ILS RWY 32 L approach. If we consider an electronic glide slope of 3 degrees, with the table of Rate-of-Descent in the IAP chart. Adjusting the angle of descent to 3 degrees and shifting the ground speed to approximately 105 knots, the estimated rate of descent is 555 ft per min.

Explanation:

The wavenumber of absorption peaks in an infrared (IR) spectrum is related to the vibrational frequencies of chemical bonds within a molecule. Different functional groups and bond types exhibit characteristic wavenumbers in the IR spectrum. When ranking carbonyl groups in a compound by increasing wavenumber, you can consider the following principles:

1. Single bonds vibrate at lower wavenumbers than double bonds.

2. Carbon-oxygen double bonds (C=O) vibrate at higher wavenumbers than carbon-oxygen single bonds (C-O).

3. The presence of electron-withdrawing groups can increase the wavenumber of the carbonyl group.

Based on these principles, here's how you can rank the carbonyl groups in the compound from lowest to highest wavenumber:

1. Carbonyl group without any adjacent electron-withdrawing groups (lowest wavenumber): This carbonyl group, if it's surrounded by alkyl or other non-electron-withdrawing groups, will have the lowest wavenumber since it's less polar and experiences weaker stretching vibrations.

2. Carbonyl group with adjacent electron-withdrawing groups: If a carbonyl group is adjacent to electron-withdrawing groups (e.g., nitro groups, fluorine atoms, etc.), it will have a higher wavenumber. The presence of these groups increases the polarity and strength of the C=O bond, causing it to vibrate at a higher frequency.

3. Carbonyl group in a conjugated system: If a carbonyl group is part of a conjugated system (alternating single and double bonds), it will have the highest wavenumber. Conjugation enhances the electron delocalization and increases the wavenumber of the carbonyl group.

So, in summary, the ranking of carbonyl groups by increasing wavenumber in an IR spectrum would generally be: carbonyl without adjacent electron-withdrawing groups < carbonyl with adjacent electron-withdrawing groups < carbonyl in a conjugated system.

The discoverer of the electron was Thomson. The model of the atom that is missing from the set is the Thomson’s model. J. J. Thomson portrayed his atom model to look like a plum pudding. In his model, he described that an atom is composed mainly of electrons.