Hello there!
To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that
a Hydrogen atom has 1 proton and 1 electron.
Protons are in the nucleus while electrons are in electron shells surrounding the nucleus.
The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:
So, the fraction of the hydrogen atom's mass that is in the nucleus is
0,9995. That means that almost all the mass of this atom is at the nucleus.
Have a nice day!
Answer:
HOMO of 1,3-butadiene and LUMO of ethylene
HOMO of ethylene LUMO of 1,3-butadiene
Explanation:
1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.
According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.
<u>Answer:</u> The chemical reaction is given below.
<u>Explanation:</u>
A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.
The reactions which occur in hydrogen-oxygen fuel cell are:
At cathode: 
At anode: 
Net reaction: 
Thus, the chemical reaction is given above.
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
Explanation:
It is known that equation for steady state concentration is as follows.
where, Q = flow rate
k = rate constant
V = volume
C = concentration of the entering air
Formula for volume of the box is as follows.
V = 
= 
= 
Now, expression to determine the discharge is as follows.
Q = Av
= 
= 0.4 
And, m (loading) = 10kg/s,
k = 0.20/hr
as 1 
L (if u want kg/L as concentration)
Now, calculate the concentration present inside as follows.
= 25 
Now, we will calculate the concentration present outwards as follows.
,
and, t = 
= 25000 s or 6.94 hr
Hence, 
= 10.47
Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is 
and 
.
