Answer:
"1.
buoyant or suspended in water or air.
2.not settled in a definite place; fluctuating or variable."
Explanation:
Hope this helps! :)
The temperature,internal energy,and kinetic energy will all increase
Explanation:
2.04 % hydrogen
32.65% sulphur
65.31% is oxygen
atomic ratio
hydrogen =2.04÷1=2.04
sulphur =32.65÷32=1.02
oxygen =65.31÷16=4.08
simplest ratio
hydrogen = 2.04÷1.02=2
sulphur =1.02÷1.02=1
oxygen =4.08÷1.02=4
empirical formula is H2SO4
<span>THE HIGHEST CONCENTRATION OF HYDROGEN IONS IS LOCATED IN THE INTER-MEMBRANE SPACE. HYDROGEN IONS REACH THE INTER-MEMBRANE SPACE THROUGH PROTEIN CHANNELS EMBEDDED IN THE MITOCHONDRIAL MEMBRANE. THE MAIN FUNCTION OF INTER MEMBRANEIS OXIDATIVE PHOSPHORLATON. ENERGY IS REQUIRED TO MOVE THE HYDROGEN IONS ACROSS THE MEMBRANE BECAUSE THE HYDROGEN IONS ARE MOVING AGAINST THE CONCENTRATION GRADIENT. H+ GOES AGAINST THE CONCENTRATION GRADIENT THE USE OF THE GRADIENT TO DRIVE ATP SYNTHASE. HYDOGEN IONS DRIVE ATP SYNTHASE IN PHTOSYNTHESIS. THIS HAPPENS WHEN HYDROGEN IONS GET PUSHED ACROSS THE MEMBRANE CREATING A HIGH CONCENTRATION INSIDE THE THYLAKOID.</span>
Answer:
(a) Rate at which 
is formed is 0.050 M/s
(b) Rate at which 
is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-
The expression for rate can be written as:-
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
Given that:- ![\frac{d[NO]}{dt}=-0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-0.050%5C%20M%2Fs)
(Negative sign shows consumption)
![-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-(-0.050\ M/s)=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%28-0.050%5C%20M%2Fs%29%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D0.050%5C%20M%2Fs)
(a) Rate at which is formed is 0.050 M/s
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.050%5C%20M%2Fs%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=0.0250\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D0.0250%5C%20M%2Fs)
(b) Rate at which is consumed is 0.0250 M/s.
