Hello there :3 Please anwser this Combined gas law problem for me . By the way just anwser #4 for me that’s all . I uploaded the
formulas on my other question if you need to use it . Please help me I don’t understand this please be an expert I will mark brianliest don’t scam me and SHOW ALL THE WORK! I just don’t understand *.* . Will report any links !
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Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:
You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
The concentrations are:
- [PCl₃]=
- [Cl₂]=
- [PCl₅]=
Replacing:
Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
The molarity of the solution is 1.69M.
<h3><u>Explanation:</u></h3>The reaction between the two reactants are:
CuNO₃+ NaI = CuI +NaNO₃.
So, 1 mole of CuNO₃ or copper nitrate produces 1 mole of CuI or Copper iodide.
Molecular weight of copper iodide = = 190.4.
So, moles of copper iodide produced here is == 0.08.
So, moles of copper nitrate that was present in the solution = 0.08.
Amount of solution used = 50ml.
So, in 50ml or solution, moles of copper nitrate present = 0.08 moles.
So, in 1 liter of solution, moles of copper nitrate present = 1.69 moles.
So, the molarity of the solution is 1.69M.