Answer: The value of the equilibrium constant Kc for this reaction is 0.088
Explanation:
where,
x = given mass
M = molar mass
= volume of solution in L
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
The given balanced equilibrium reaction is,
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CaO]\times [CO_2]}{[CaCO_3]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCaO%5D%5Ctimes%20%5BCO_2%5D%7D%7B%5BCaCO_3%5D%7D)
Now put all the given values in this expression, we get :
Answer:
22 in total but round it up to 23
Answer:
A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.
Explanation:
A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.
1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J
ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J
B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂
= -4.3J -(-6.0 J)= 1.7 J. Since U₁ - U₂ > 0 , U₂ < U₁ , so the second system's internal energy increase less and is lower in energy and is more stable.
Missing question: <span>A 5.00 L sample of O2 at a given temperature and pressure contains a 1.08x10^23 molecules. How many molecules would be contained in each of the following at the same temperature and pressure? </span>
a) 5.00 L H2.
<span>b) 5.00 L CO2.
Use </span>Avogadro's Law: The Volume Amount Law: <span>equal </span>volumes<span> of all gases, at the same temperature and pressure, have the same </span>number<span> of molecules. Because hydrogen and carbon(IV) oxide are gases, number of molecules are the same as number of oxygen molecules, so:
a) N(H</span>₂) = 1.08·10²³.
b) N(CO₂) = 1.08·10²³
