Explanation:
For reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.
The given balanced equation has been:
\rm O_2\;+\;2\;H_2\;\rightarrow\;H_2OO2+2H2→H2O
From the equation, 1 mole of oxygen reacts with 2 mole of hydrogen to give 1 mole of water.
The mass of oxygen has been: 8.75 g,
Moles = \rm \dfrac{weight}{molecular\;weight}molecularweightweight
Moles of oxygen = \rm \dfrac{8.75}{32}328.75
Moles of oxygen = 0.27 mol
Since,
1 mole Oxygen = 2 mole hydrogen
0.21 mol oxygen = 0.54 mol hydrogen
Mass of hydrogen = moles \times× molecular weight
Mass of hydrogen = 0.54 \times× 2
Mass of hydrogen = 1.08 grams.
Thus, for reacting with 8.75 grams of oxygen, 1.08 grams of hydrogen is required.
Answer: An electron will jump to a higher energy level when excited by an external energy gain such as a large heat increase or the presence of an electrical field, or collision with another electron.
Explanation:
If we're talking about the amount of electron rings and the dots within the rings around an atom, the amount will be 1 ring, 2 dots.
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)
For a neutral solution,
[H+][OH-] = 1 x 10⁻¹⁴