Answer:
K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2
Explanation:
The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.
The reaction above is a reversible reaction and the law of mass action also applies to it.
The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2
Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]
Then at equilibrium,
r1 = r2
k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K
where K is the equilibrium constant for the reaction.
Answer:
Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km
<em>Note: The question is missing some parts. The complete question is as follows;</em>
<em>A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]</em>
Explanation:
Avogadro number = 6.02 * 10²³
thickness of a penny = 1.0 mm
I mm = 0.001 m
Thickness of Avogadro number of pennies stacked one upon another will be:
6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m
Distance in km;
1 m = 0.001 km
therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km
Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km
Answer: The final temperature of the system will be 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of steam = 25 g
= mass of water = 0.2384 kg = 238.4 g (1kg=1000g)
= final temperature = ?
= temperature of steam = 
= temperature of water = 
= specific heat of steam = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![25g\times 1.996J/g^0C\times (T_{final}-116)=-[238.4g\times 4.184J/g^0C\times (T_{final}-8)]](https://tex.z-dn.net/?f=25g%5Ctimes%201.996J%2Fg%5E0C%5Ctimes%20%28T_%7Bfinal%7D-116%29%3D-%5B238.4g%5Ctimes%204.184J%2Fg%5E0C%5Ctimes%20%28T_%7Bfinal%7D-8%29%5D)
Therefore, the final temperature of the system will be
