Explanation:
It is known that the specific heat capacity of Liver
is 3.59 kJ
It is given that :
Initial temperature of Liver = Body temperature =
= 310 K
Final temperature of Liver = 180 K
Relation between heat energy, mass, and change in temperature is as follows.
Q =
Now, putting the given values into the above formula as follows.
Q = 
Q =
= 700.05 kJ
Therefore, we can conclude that amount of heat which must be removed from the liver is 700.05 kJ.
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)
Answer:
2.61 g of NO will be formed
The limiting reagent is the O₂
Explanation:
The reaction is:
4NH₃ + 5O₂ → 4NO + 6H₂O
We convert the mass of the reactants to moles:
3.25g / 17 g/mol = 0.191 moles of NH₃
3.50g / 32 g/mol =0.109 moles of O₂
Let's determine the limiting reactant by stoichiometry:
4 moles of ammonia react with 5 moles of oxygen
Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent
Ratio with NO is 5:4
5 moles of oxygen produce 4 moles of NO
0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO
We convert the moles to mass, to get the answer
0.0872 mol . 30g / 1 mol = 2.61 g
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