Answer:
So the answer would be 10 moles
Explanation:
1) Start with the molecular formula for water: 
2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.
(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of 
)
The mole ratio is 1 moles of 
to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.
3) that means if 10 moles of water decompose, they decompose into 10 moles of 
and 10 moles of O.
Extra:
About what I was saying before about the question being slightly interesting:
10 moles of pure oxygen is produced but free state oxygen exists as so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).
I hope this helps!
Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
Answer:
none of the above
Explanation:
A system is said to have attained dynamic equilibrium when the forward and reverse reactions proceed at the same rate. That is;
Rate of forward reaction = Rate of reverse reaction
The implication of this is that the concentrations of reactants and products remain constant when dynamic equilibrium is attained in a system. This does not mean that the reactant and product concentrations become equal; it rather means that their concentrations do not significantly change once dynamic equilibrium has been attained.
Iodine electron configuration is:
1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.
So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5
