If the substance mixes with water it's polar. If it doesn't it ms non polar.
Answer:
The pressure inside the container will be 3.3 atmospheres
Explanation:
The relationship between the temperature and pressure of a gas occupying a fixed volume is given by Gay-Lussac's law which states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is kept constant.
Mathematically, it expressed as: P₁/T₁ = P₂/T₂
where P₁ is initial pressure, T₁ is initial temperature, P₂ is final pressure, T₂ is final temperature.
The above expression shows that the ratio of the pressure and temperature is always constant.
In the given question, the gas in the can attains the temperature of its environment.
P₁ = 3 atm,
T₁ = 25 °C = (273.15 + 25) K = 298.15 K,
P₂ = ?
T₂ = (55 °C = 273.15 + 55) K = 328.15 K
Substituting the values in the equation
3/298.15 = P₂/328.15
P₂ = 3 × 328.15/298.15
P₂ = 3.3 atm
Therefore, the pressure inside the container will be 3.3 atmospheres
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
Answer:
Option B. The reaction will shift to the left in the direction of the reactants.
Explanation:
The equation for the reaction is given below:
CO₂ + 2H₂O <=> CH₄ + O₂
Enthalpy change (ΔH) = +890 KJ
The reaction illustrated by the equation is endothermic reaction since the enthalpy change (ΔH) is positive.
Increasing the temperature of an endothermic reaction will shift the equilibrium position to the right and decrease the temperature will shift the equilibrium position to the left.
Therefore, decreasing the temperature of the system illustrated by the equation above, will shift the reaction to the left in the direction of the reactants.
Thus, option B gives the right answer to the question.
