<h2>a)
The rate at which 
is formed is 0.066 M/s</h2><h2>b)
The rate at which molecular oxygen 
is reacting is 0.033 M/s</h2>
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of 
= ![-\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
= 0.066 M/s
Rate in terms of disappearance of = ![-\frac{1d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D)
Rate in terms of appearance of = ![\frac{1d[NO_2]}{2dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7B2dt%7D)
1. The rate of formation of
![-\frac{d[NO_2]}{2dt}=\frac{1d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO_2%5D%7D%7B2dt%7D%3D%5Cfrac%7B1d%5BNO%5D%7D%7B2dt%7D)
![\frac{1d[NO_2]}{dt}=\frac{2}{2}\times 0.066M/s=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7B1d%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.066M%2Fs)
2. The rate of disappearance of
![-\frac{1d[O_2]}{dt}=\frac{d[NO]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7B2dt%7D)
![-\frac{1d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
Learn more about rate law
brainly.com/question/13019661
https://brainly.in/question/1297322
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
CH3NH2 can only have as many hydrogen bonds as hydrogen bonding sites in the molecule. CH3NH2 has two N−H bonds and a lone pair of electrons on the nitrogen atom. Therefore, CH3NH2 can form three hydrogen bonds with water.
<h2>
Answer: 4.72 × 10⁷ s⁻¹ or Hz</h2><h3>
Explanation: </h3>
frequency = speed of light ÷ wavelength
frequency = (2.99 × 10⁸ m/s) ÷ (6.34 × 10⁸ m)
= 4.72 × 10⁷ s⁻¹ or Hz
Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
= 4.19 
= 1.9
Heat of vaporization (
) at 1 atm and 
is 2259 kJ/kg
= 0
Therefore, calculate the enthalpy of water vapor at 1 atm and as follows.
= +
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given 
and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and is calculated as follows.
= 
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and is calculated as follows.
= 
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and as follows.
= 
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and latent heat of vaporization is 2076 kJ/kg.
