It's 1. Melting a substance. The rest are chemical changes
Answer :
(a) The rate of 
formed is, 0.066 M/s
(b) The rate of 
formed is, 0.033 M/s
Explanation : Given,
![\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
= 0.066 M/s
The balanced chemical reaction is,
The rate of disappearance of 
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of = ![-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The rate of formation of = ![\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
As we know that,
= 0.066 M/s
(a) Now we have to determine the rate of formed.
![\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D0.066M%2Fs)
The rate of formed is, 0.066 M/s
(b) Now we have to determine the rate of molecular oxygen reacting.
![-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
The rate of formed is, 0.033 M/s
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:
brainly.com/question/14735801
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This is valid only if the two elements have the same valence
