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)
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Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
<u />
<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
