Solute particles can be atoms, ions or molecules.
Explanation:
Solute is the material which has to be mixed in the solvent to prepare a solution. So the concentration of solute should be less than the solvent. Also the solute and solvent should be of same nature other they will not dissolve with each other. The solute can be made up of atoms, ions or molecules depending upon the solvent. If the solvent concentration is in moles, then the solute concentration can be taken as atoms, ions or molecules. Also the saturation point plays a main role in deciding the kind of particles taken for the solute.
Oxygen, fluorine and iodine are diatomic elements. Flourine is more reactive than the other two because it is the closest away to filling its outer layer of electrons and becoming stable like a noble gas.
At room temperature, O2 is in gaseous state.
a gas has no definite volume or definite shape. It occupies volume of container and attains shape of container only.
Thus
It has no definite volume and takes the shape of its container.
Its particles move fast enough to overcome the attraction between them.: the gas molecules have minimum intermolecular interactions and have high kinetic energy.
It has more energy than it would at a cooler temperature: the kinetic energy of gas molecules increases with increase in temperature. Thus the energy increases with temperature and decreases with decrease in temperature.
Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar
Answer:
0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent
Explanation:
The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.
