<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.
Answer:
The carrying capacity of this population would be 125 we know this because we see that this number occur multiple times and seems to be the tipping point after which the number of the population always go down
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Above it says the molecular weights are
NH3- 17g/mol and SF6-146 g/mol
Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.
<span><span><span>= 1mole / </span><span>146.048g *</span></span> 0.85g</span>
so we would need 0.00582000438 moles of NH3 to have the same number of molecules.
One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:
<span><span><span>= 17.030519999989988 g / </span><span>mole * </span></span>0.00582000438moles</span>
that equals 0.09911770099 grams.
so 0.09911770099 grams is the answer if you round that you get about 0.1 grams