Answer:
The wavelength of sunlight is required to break the bond in one oxygen molecule is 242 nm.
Explanation:
Energy required to break the 1 mol oxygen-oxygen bond:
=495 kJ=495000J
1 mol = 
molecules
Energy required to break 1 molecule of oxygen molecule= E
= 
The energy required to break the 1 molecule of oxygen is equal to enrgy of one photon of a sunlight.
where,
= wavelength of the light
E = energy of the photon
h = Planck's constant = 
c = speed of light = 
The wavelength of sunlight is required to break the bond in one oxygen molecule is 242 nm.
The compound that ontains the most deactivated aromatic ring is the one called Nitrogen dioxide also knwon as NO2. Let me explain a little bit more about the aromatic rings. These rings are simple aromatics, or arenes, that consist of a conjugated planar ring system. Nitrogen generally is a simple aromatic ring that can be heterocyclic because containsnon-carbon ring atoms. The rings of the nitrogen can be separated into basic aromatic rings and that is why this ompound is also refered as a deactivated aromatic ring.
Answer:
a. 1/3
In the picture- D. 4
Explanation:
The product of the gradients of perpendicular lines =-1
m1m2=-1
Given an equation of a line in the form y=mx+c, m is the gradient of the line.
For y= -3x-9, m=-3
m1m2=-1
-3×m2=-1
m2=1/3
The gradient of the perpendicular line is 1/3
For the question in the picture, y=-1/4x-9, m=-1/4
m1×m2=-1
-1/4m2=-1
m2=4
The slope of the line perpendicular to y=-1/4x-9 is 4
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
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Answer:
Continuously recurring, or cycling maybe?
