Answer:
one of the atoms sharing electrons is more electronegative than the other atom
Explanation:
Answer:
Explanation:
From the information given:


no of moles of
= 0.01 L × 0.0010 mol/L
no of moles of
= 
no of moles of
= 0.01 L × 0.00010 mol/L
no of moles of
= 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
Answer:
Here is your answer mate :D
The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :

NO :

Cl2 :

![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)