Answer:
Mass is lost due to the conversion of mass to energy
Explanation:
The question is not complete, the complete question is given as:
⇒ 
total mass equals 236.053 u total mass equals 235.868 u
Which statement explains the energy term in this reaction? (1) Mass is gained due to the conversion of mass to energy. (2) Mass is gained due to the conversion of energy to mass. (3) Mass is lost due to the conversion of mass to energy. (4) Mass is lost due to the conversion of energy to mass.
Answer: From Einstein’s equation E = mc², when a radioisotope element undergoes fission or fusion in a nuclear reaction, it loses a tiny amount of mass.This mass lost is converted to energy.
The law of conservation of energy holds for this type of reaction (i.e the sum of mass and energy is remains the same in a nuclear reaction). Mass changes to energy, but the total amount of mass and energy combined remains the same before and after a nuclear reaction.
From the reaction above, the total decrease in mass = 236.053 - 235.868 = 0.185 u
Molecular weight of N2
gram Atomic weight of N is 14 g.
gram Molecular weight = 14 g×2
= 28 g
Liter concepts
1 liter of gas always occupies 1 gram molecular weight .
Application of the concept
5.75 liter gives 0.257g
1 mole will occupy 28 g
0.257 will occupy 28 g × 0.257
= 7.196 g
The mass of the gas is 7.196 g.
Hope it helps you
TMS stands for Tetramethylsilane
Formula : Si(CH₃)₄
It is used as internal standard for chemical shift
In TMS all the four methyl groups are similar thus all the 12 hydrogen are chemically equivalent. They will show proton NMR at same level and which will appear as a singlet.
We have defined this singlet to be as reference set as zero.
Generally the other organic compounds show chemical shift downfield.So have positive chemical shift with respect to TMS shift.
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
![rate = k*[O_{3}][NO]](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D%5BNO%5D%20)
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
![rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D_%7B0%7D%5BNO%5D_%7B0%7D%20%3D%203.91%20%5Ccdot%2010%5E%7B6%7D%20M%5E%7B-1%7Ds%5E%7B-1%7D%2A2.35%5Ccdot%2010%5E%7B-6%7D%20M%2A7.74%20%5Ccdot%2010%5E%7B-5%7D%20M%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%20)
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
