Answer:
more energy
Explanation:
shorter lengths in bonds require more energy because atoms are strongly connected in short length bonds
2[H⁺] + [O²⁻] → H₂O
Explanation:
Half reactions are usually composed of components of oxidation-reduction (redox) reaction where one element is losing electrons while another is gaining so they can come together into a compound. The half-reaction shows part of a reaction where the combining elements are charged and the product, from the redox, is neutral.
In this case, the H⁺ is carrying the positive charge because it is donating an electron to O²⁻, which is the reason the O²⁻ is carrying the negative charge. Note that we need 2 H⁺ for every one O²⁻. The O is being reduced while the H is being oxidized.
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For more on half-reaction check out;
brainly.com/question/7570973
brainly.com/question/11577402
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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
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Balanced equation brainly.com/question/9325293
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The are made from star stuffs...
Galaxies are composed of stars, dust, and dark matter, all held together by gravity. Below we discuss galaxy formation, galactic collisions and other facts about these so-called “island universes.” The Milky Way Galaxy is organized into spiral arms of giant stars that illuminate interstellar gas and dust.
Hope that helps........ (◕‿◕✿)
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
