Answer:
Explanation:
Given that:
From above:
To predict the effect of the addition of Br₂(g);
The addition of Br₂(g) will favor the equilibrium to shift to the left i.e. formation of NOBr
The removal of some NOBr will cause the equilibrium position to shift to the left side. This is because concentration on the left side is decreased and the concentration on the right side will be increased. Thus, the equilibrium will shift towards where the concentration is reduced which is the left side.
This question is incomplete because the options are missing; here are the options:
Which of the following is LESS dense than water?
The spoon
The glass
The tablets
The bubbles
The correct answer to this question is The bubbles
Explanation:
In general, the density of materials and substances affects their buoyancy. This implies in water less dense materials will float and those with higher density will sink. In the situation presented, the only element that is less dense than water are bubbles; this is shown by the movement of the bubbles as these originate in the bottom of the glass of water but they rise to the surface, which shows they are less dense than water.
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
C. a scientific approach to answering questions
