Answer:
<em>When salt is dissolved in water</em>, many physical properties change, among them the so called colligative properties:
- The vapor pressure of water decreases,
- The boiling point increases,
- The freezing point decreases, and
- Osmotic pressure appears.
Explanation:
Colligative properties are the physical properties of the solvents whose change is determined by the number of particles (moles or ions) of the solute added.
The colligative properties are: vapor pressure, boiling point, freezing point, and osmotic pressure.
<u>Vapor pressure</u>:
The vapor pressure is the pressure exerted by the vapor of a lquid over its surface, in a closed vessel.
The vapor pressure increases when a solute is added, because the presence of the solute causes less solvent molecules to be near the surface ready to escape to the vapor phase, which means that the vapor pressure is lower.
<u>Boiling point</u>:
The boiling point is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure. Since we have seen that the vapor pressure of water decreases when a solute occupies part of the surface, now more temperature will be required for the water molecules reach the atmospheric pressure. So, the boiling point increases when salt is dissolved in water.
<u>Freezing point</u>:
The freezing point is the temperarute at which the vapor pressure of the liquid and the solid are equal. Since, the vapor pressure of water with salt is lower than that of the pure water, the vapor pressure of the liquid and solid with salt will be equal at a lower temperature. Hence, the freezing point is lower (decreases).
<u>Osmotic pressure</u>:
Osmotic pressure is the additional pressure that must be exerted over a solution to make that the vapor pressure of the solvent in the solution equals the vapor pressure of the pure solvent. This additional pressure is proportional to the concentration of the solute: the higher the salt concentration the higher the osmotic pressure.
Answer:
The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%
Explanation:
Here we have
Propane gas with molecular formula C₃H₈, molar mass = 44.1 g/mol combining with O₂ as follows
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Therefore, 1 mole of C₃H₈ combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O
Mass of propane = 0.1240 kg = 124.0 g
Number of moles of propane = mass of propane/(molar mass of propane)
The number of moles of propane = 124/44.1 = 2.812 moles
The molar mass of CO₂ = 44.01 g/mol
Mass of CO₂ = 0.3110 kg = 311.0 g
Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)
The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles
Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂
Therefore;
The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess
Hence
The percentage yield = 83.77%.
Hello!
The hybridization of the C atom in CH₂Br₂ is sp3
When bonding, the orbitals "s" and "p" from C atoms interact to form hybridized orbitals. If the C atom has 4 sigma bonds, as is the case in CH₂Br₂, there are 4 hybridized orbitals required, so 1 "s" orbital and 3 "p" orbitals hybridize to form an sp3 hybrid orbital. This orbital has tetrahedral geometry and the bond angle is 109,5 °.
Have a nice day!
Answer:
=342g
Explanation:
atomic mass of C = 12g
atomic mass of H = 1g
atomic mass of O = 16g
Solution;
C12 H22 O11
= 12 (12) + 22 (1) + 11(16)
= 144+ 22 + 176
= 342g
The answer is B have a good day
