0 Kelvin is the temperature where the is no kinetic energy, so the answer is 0.
Hey there!:
Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :
MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)
Write the total ionic equation for the reaction :
Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)
Therefore:
Cancel the spectator ions on both sides:
Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)
Hope that helps!
Answer:
Partial pressure of ICl at equilibrium is 1.69 atm
Partial pressure of chloride gas at equilibrium is 0.00631 atm.
Partial pressure of iodine gas at equilibrium is 0.00631 atm.
Explanation:
The partial pressure of ICl initially = 1.70 atm
The equilibrium constant of the reaction = ![K_p=1.40\times 10^{-5}](https://tex.z-dn.net/?f=K_p%3D1.40%5Ctimes%2010%5E%7B-5%7D)
![2ICl(g)\rightequilibrium Cl_2(g)+I_2(g)](https://tex.z-dn.net/?f=2ICl%28g%29%5Crightequilibrium%20Cl_2%28g%29%2BI_2%28g%29)
Initially
1.70 atm 0 0
At equilibrium
(1.70-2p) p p
The expression of equilibrium constant is given by :
![K_p=\frac{p_{Cl_2}\times p_{I_2}}{(p_{ICl})^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_%7BCl_2%7D%5Ctimes%20p_%7BI_2%7D%7D%7B%28p_%7BICl%7D%29%5E2%7D)
![1.40\times 10^{-5}=\frac{p\times p}{(1.70-2p)^2}](https://tex.z-dn.net/?f=1.40%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7Bp%5Ctimes%20p%7D%7B%281.70-2p%29%5E2%7D)
Solving for p :
p = 0.00631 atm
Partial pressure of ICl at equilibrium = 1.70 atm- 2 0.00631 atm = 1.69 atm
Partial pressure of chloride gas at equilibrium = 0.00631 atm.
Partial pressure of iodine gas at equilibrium = 0.00631 atm.
Explanation:
here's the answer to your question about
Answer:
Because they convert the electrical voltage produced by increasing and then it goes into the power lines
Explanation: