The true statements are;
<h3>What is a redox reaction?</h3>
We define a redox reaction as one in which a specie is oxidized and another is reduced.
Now;
Eo cell = cell potential = -0.13 V - (+0.34 V) = -0.47 V
n =number of moles of electrons = 2 mole of electrons
K = equilibrium constant
ΔG = change in free energy
Eo cell = 0.0592/n log K
-0.47 = 0.0592/2 log K
log K = -0.47 * 2/0.0592
K = 1.3 * 10^-16
ΔG = -nFEo cell
ΔG = -(2 * 96500 * -0.47)
ΔG = 90.7kJ
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they all have one thing in common and that its all made up of atoms. When these components are active it creates energy
Solid
, Inorganic,
Naturally Orcurring,
Defintite
Chemical Compostion,
Definite Crystalline Structure
5 Physical Properties
Hardness,
Color,
Crystal Shape,
Streak
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that 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>
<em />
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
<em></em>
<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
<em></em>
<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
<em></em>
<em>C. at low temperature and low pressure.</em>
<em></em>
Answer:
75 g
Step-by-step explanation:
<em>Step 1. </em>Calculate the molar mass of CaCO₃
Ca = 40.08
C = 12.01
3O = 3 × 16.00 = <u> 48.00
</u>
Total = 100.09 g/mol
<em>Step 2</em>. Calculate the mass of the pellet
Mass of CaCO₃ = 0.75 mol × (100.09 g/1 mol)
Mass of CaCO₃ = 75 g
