Answer:
The number of moles of the gas is 9.295 moles or 9.30 moles
Explanation:
We use PV = nRT
Where P = 4.87 atm;
V = 67.54 L
R= 0.0821Latm/molK
T = 158 C = 158 +273 K = 431 K
the number of moles can be obtained by substituting the values in the respective columns and solve for n
n = PV / RT
n = 4.87 * 67.54 / 0.0821 * 431
n = 328.9198 / 35.3851
n = 9.295moles
The number of moles is approximately 9.30moles.
please check the attached image, i have given this answer b4
<span>Research shows that fabric softeners decrease flame resistance. The flame resistance of a fabric can be increased in a variety of ways. Most often, the fabric is treated with special flame resistant chemicals. Synthetic fabrics, such as polyester, are naturally flame resistant. These fabrics have a property called thermoplasticity, where the fabric naturally melts and shrinks when exposed to flame. This prevents these fabrics from igniting and burning as easily as other fabrics. Basic properties of a fabric, such as porosity, can also influence flame resistance. Larger pores allow for oxygen to be more present in the fabric, making it more flammable.</span>
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
This set up of a conversion table should show you that if you multiply
the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>
