Answer:
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Explanation:
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The balanced chemical reaction:
K2SO4 + O2 = 2KO2 + SO2
Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:
7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,

initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)


as
is very small then we can neglect 
therefore ,



Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
<u>Answer:</u> The equilibrium concentration of
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:

The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:

![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of
is 1.285 M.