Answer:
The rate of forward reaction increases.
Explanation:
- Le Châtelier's principle states that <em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
Increasing the volume of oxygen:
will increase the concentration of oxygen (reactants), so the equilibrium will be shifted to the right side (products side) to suppress the effect of increasing the volume of oxygen.
<em>So, the right choice is: The rate of forward reaction increases.
</em>
<em></em>
Join or be joined securely to something else, typically by means of an adhesive substance, heat, or pressure.
Answer: 0.02442 with 5 s.f. Explanation: (23.4g +25.2g +24.3g + 24.4g +24.8g)/5 = 24.42g. 1000g/kg so 24.42g = 0.02442 kg
Answer:
D). She should use a bar graph. The bottom of each bar should be labeled with the category of garbage. The vertical side should have a scale and be labeled "kilograms per year."
Explanation:
Bar graphs are characterized as the graphic tool which are primarily employed to establish comparison between various groups or to observe the changes occurred over time.
As per the question, the environmental scientist must opt for the 'bar graph' as it will assist her in displaying the data of various categories(paper, yard waste, metals, etc.) through bars on the x-axis while the vertical side or y-axis would show 'kilograms per year' which will display the differences in their amounts. Thus, <u>option D</u> is the correct answer.
Answer:
337.22 K
Explanation:
Given that:
P₁ = 1 atm
T₁ = 350 K
P₂ = 0.639 atm
T₂ = ??? (unknown)
R(rate constant) = 8.34 J k⁻¹ mol⁻¹
Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.
Clausius-Clapeyron equation can be written as:
![In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]](https://tex.z-dn.net/?f=In%5Cfrac%7BP_2%7D%7BP_1%7D%3D%5Cfrac%7B%5Cdelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7BT_2-T_1%7D%7BT_2T_1%7D%5D)
Substituting our values given; we have:
![In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]](https://tex.z-dn.net/?f=In%5Cfrac%7B0.639%7D%7B1%7D%3D%28%5Cfrac%7B34.4%2A10%5E3J%2Fmol%7D%7B8.314%20J%20K%5E%7B-1%7Dmol%5E%7B-1%7D%7D%29%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%5D)
![In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]](https://tex.z-dn.net/?f=In%28%7B0.639%7D%29%3D%28%5Cfrac%7B34.4%2A10%5E3%7D%7B8.314K%5E%7B-1%7D%7D%29%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%5D)
![- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K](https://tex.z-dn.net/?f=-%200.4479%20%3D%2041317.599%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5DK)
![-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]](https://tex.z-dn.net/?f=-%5Cfrac%7B0.4479%7D%7B4137.599%7D%20%3D%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5D)
![- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]](https://tex.z-dn.net/?f=-%201.0825118%2A10%5E%7B-4%7D%20%3D%20%5B%5Cfrac%7BT_2-350%7D%7B350T_2%7D%20%5D)
∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is <u>337.22</u> K.
