Answer:
Heyo (Ish Mash Potato) XD
Explanation:
a freighter carrying a cargo of uranium hexafluoride sank in the english channel in 1984. the cargo of uranium hexafluoride weighed 2.25x10 to the eight power. kg and was contained in 30 drums, each having a volume of 1.62x10 to the sixth power L. what is the density in g/ml, of uranium hexafloride.
Hope Mr. Mash Potato Helped!!!
Answer:
The reaction mentioned in the question is a decomposition reaction.
Explanation:
Decomposition reaction is a specific type of chemical reaction in which the reactant molecule undergo break down to generate its respective products.
NaCl ⇒ Na + Cl2
In the above reaction sodium chloride(NaCl) undergo decomposition or break down to generate Sodium(Na) and chlorine gas(Cl2).
<u>Given:</u>
Enthalpy change (ΔH) for SO3 decomposition = +790 kJ
Moles of SO3 = 2.1 moles
<u>To determine:</u>
Energy required when 2.1 moles of SO3 reacts
<u>Explanation:</u>
The decomposition reaction is -
2SO3(g) → 2S(s) + 3O2 (g)
Energy required when 2 moles of SO3 reacts is 790 kJ
Thus, for 2.1 moles of SO3 the energy requirement would be
= 2.1 moles SO3 * 790 kJ/2 moles SO3 = 829.5 kJ
Ans: 830 kJ are required when 2.1 moles of SO3 reacts.
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)
