Not sure if this is correct but i would choose the steel tray.
Answer:
Reaction 1: Kc increases
Reaction 2: Kc decreases
Reaction 3: The is no change
Explanation:
Let us consider the following reactions:
Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol
Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol
Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol
To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>
Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.
Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.
Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.
First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl
since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol
1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.
8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂
2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl
3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl
3) lastly, we convert the moles of HCl to grams using the molar mass.
molar mass of HCl= 1.01 + 35.5= 36.51 g/mol
7.1x10⁻⁴ mol HCl (36.51 g/mol)=<span> 0.026 grams HCl</span>
Answer:
Explanation:
The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below
H
|
H₂N - C - C =O
| \
H OH
The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.
One way of knowing that oxygen was the gas removed from the volume of air and not another is to know what the volume of air is made of first. When the composition of the volume of air is already identified, then next would be the process of separating these elements from each other and as to which is to be separated first. This would usually lead to knowing their masses, their boiling and freezing points, the temperatures at which they condense, and so on. This is to identify their differences to each other and use those differences to successfully separate those elements to each other.
