A substance can dissolve in another when they have thee same type of intermolecular interaction.
<h3>What is solubility?</h3>
The term solubility of a solute refers to the extent to which a solute dissolve in a solvent. We must know that a substance can dissolve in another when they have thee same type of intermolecular interaction.
Thus;
a) Octane (C8H18) mixes well with CCl4 because they are both non polar substances.
b) Methanol (CH3OH) is mixed with water in all ratios because the both are polar substances.
c) NaBr dissolves very poorly in acetone (CH3 ― CO ― CH3) because acetone is only slightly polar.
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Answer:
The standard enthalpy change for the reaction at 
is -2043.999kJ
Explanation:
Standard enthalpy change (
) for the given reaction is expressed as:
![\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
Where 
refers standard enthalpy of formation
Plug in all the given values from literature in the above equation:
![\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5B3mol%5Ctimes%20%28-393.509kJ%2Fmol%29%5D%2B%5B4mol%5Ctimes%20%28-241.818kJ%2Fmol%29%5D-%5B1mol%5Ctimes%20%28-103.8kJ%2Fmol%29%5D-%5B5mol%5Ctimes%20%280kJ%2Fmol%29%5D%3D-2043.999kJ)
Answer: Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behavior of real gases. The term ideal gas refers to a hypothetical gas composed of molecules which follow a few rules:
Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container. [What is an elastic collision?]
Ideal gas molecules themselves take up no volume. The gas takes up volume since the molecules expand into a large region of space, but the Ideal gas molecules are approximated as point particles that have no volume in and of themselves.
If this sounds too ideal to be true, you're right. There are no gases that are exactly ideal, but there are plenty of gases that are close enough that the concept of an ideal gas is an extremely useful approximation for many situations. In fact, for temperatures near room temperature and pressures near atmospheric pressure, many of the gases we care about are very nearly ideal.
If the pressure of the gas is too large (e.g. hundreds of times larger than atmospheric pressure), or the temperature is too low (e.g.
−
200
C
−200 Cminus, 200, start text, space, C, end text) there can be significant deviations from the ideal gas law.
Explanation:
Answer:
619°C
Explanation:
Given data:
Initial volume of gas = 736 mL
Initial temperature = 15.0°C
Final volume of gas = 2.28 L
Final temperature = ?
Solution:
Initial volume of gas = 736 mL (736mL× 1L/1000 mL = 0.736 L)
Initial temperature = 15.0°C (15+273 = 288 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂/V₁
T₂ = 2.28 L × 288 K / 0.736 L
T₂ = 656.6 L.K / 0.736 L
T₂ = 892.2 K
K to °C:
892.2 - 273.15 = 619°C
