Answer: Aluminum reacts with oxygen to produce aluminum oxide as follows: 4Al + 3O2 → 2Al2O3 a.
Explanation: I really really <u>hope</u> that was <u>helpful.</u>
Use the PV = nRT equation T is in Kelvins = 31 + 273 = 304 K
P(0.5) = (2.91)(0.0821)(304)
P(0.5) = 72.6289
P = 145.25 atm or 1.45x10^2 atm
6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
Explanation:
Given equation;
NaC₂H₃O₂ + Fe₂O₃ → Fe(C₂H₃O₂)₃ + Na₂O
To find the coefficient that will balance this we equation, let us set up simple mathematical algebraic expressions that we can readily solve.
Let us have at the back of our mind that, in every chemical reaction, the number of atom is usually conserved.
aNaC₂H₃O₂ + bFe₂O₃ → cFe(C₂H₃O₂)₃ + dNa₂O
a, b, c and d are the coefficients that will balance the equation.
conserving Na; a = 2d
C: 2a = 6c
H: 3a = 9c
O; 2a + 3b = 6c + d
Fe: 2b = c
let a = 1
solving:
2a = 6c
2(1) = 6c
c = 
2b = c
b = 
= 
d = 2a + 3b - 6c = 2(1 ) + (3 x ) - (6 x 
) = 
Now multiply through by 6
a = 6, b = 1, c = 2 and d = 3
6NaC₂H₃O₂ + Fe₂O₃ → 2Fe(C₂H₃O₂)₃ + 3Na₂O
learn more:
Balanced equation brainly.com/question/9325293
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Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Your answer is correct.
MgO(s) + H2O(l) ----> Mg(OH)2(aq)
:-) ;-)
