<span>Separate this redox reaction into its component half-reactions.
Cl2 + 2Na ----> 2NaCl
reduction: Cl2 + 2 e- ----> 2Cl-1
oxidation: 2Na ----> 2Na+ & 2 e-
2) Write a balanced overall reaction from these unbalanced half-reactions:
oxidation: Sn ----> Sn^2+ & 2 e-
reduction: 2Ag^+ & 2e- ----> 2Ag
giving us
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
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When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
Greater than 23mcg/dl is considered a high level of morning cortisol.
Normally, cortisol levels rise during the early morning hours and are highest about 7AM. They drop very low in the evening and during the early phase of sleep. If you do not have this daily change (diurnal rhythm) in cortisol levels, you may have overactive adrenal glands. This condition is called Cushing's syndrome.
<u>Answer:</u> Increasing temperature
<u>Explanation:</u>
The Principle of Le Chatelier states that <u>if a system in equilibrium is subjected to a change of conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.
</u>
The variation of one or several of the following factors can alter the equilibrium condition in a chemical reaction:
- Temperature
- The pressure
- The volume
- The concentration of reactants or products
In the case of the reaction in the question, <u>the change that moves the balance to the left will be the one that moves it towards the reagents</u>, that is, that favors the production of reagents instead of products.
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Decreasing the concentration of SO3 and increasing the concentration of SO2 <u>will favor the production of SO3</u>, which is the product of the reaction.
- Decreasing the volume increases the pressure of the system and the balance will move to where there is less number of moles. In the case of the reaction in question, we have 3 moles of molecules in the reactants (1 mole of O2 + 2 moles of SO2) while in the products there are 2 moles of SO3 only, therefore, <u>decreasing the volume will displace the balance to the right</u>, which corresponds to the sense in which there is less number of moles.
The reaction of the question is an exothermic since ΔH <0, therefore in the reaction heat is produced and it can be written in the following way,
2SO2(g) + O2(g) ⇌ 2SO3(g) + heat
- So, if we increase the temperature we will be adding heat to the system, so the balance would move to the left to compensate for the excess heat in the system.
