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Sagot :
To determine which substance cancels out when combining the given intermediate chemical equations, let's first write out the equations clearly:
1. [tex]\( 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \)[/tex]
2. [tex]\( PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \)[/tex]
Let's examine these two equations. Notably, we must identify if any substance appears as both a product in one equation and a reactant in another, as this is the substance that will cancel out.
1. The first equation produces [tex]\( 2PCl_3(l) \)[/tex] in the product side.
2. The second equation uses [tex]\( PCl_3(l) \)[/tex] as a reactant.
So, we observe that [tex]\( PCl_3 \)[/tex] appears as a product in the first equation and as a reactant in the second equation. Therefore, when we combine these two equations, [tex]\( PCl_3 \)[/tex] will cancel out.
Thus, the substance that cancels out when combining these intermediate chemical equations is [tex]\( PCl_3 \)[/tex].
So, the answer is:
[tex]\[ \boxed{PCl_3} \][/tex]
1. [tex]\( 2P(s) + 3Cl_2(g) \rightarrow 2PCl_3(l) \)[/tex]
2. [tex]\( PCl_3(l) + Cl_2(g) \rightarrow PCl_5(s) \)[/tex]
Let's examine these two equations. Notably, we must identify if any substance appears as both a product in one equation and a reactant in another, as this is the substance that will cancel out.
1. The first equation produces [tex]\( 2PCl_3(l) \)[/tex] in the product side.
2. The second equation uses [tex]\( PCl_3(l) \)[/tex] as a reactant.
So, we observe that [tex]\( PCl_3 \)[/tex] appears as a product in the first equation and as a reactant in the second equation. Therefore, when we combine these two equations, [tex]\( PCl_3 \)[/tex] will cancel out.
Thus, the substance that cancels out when combining these intermediate chemical equations is [tex]\( PCl_3 \)[/tex].
So, the answer is:
[tex]\[ \boxed{PCl_3} \][/tex]
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