Answer:
4,2,1,3
Explanation:
4. A small ribosomal subunit binds to a molecule of mRNA5. the anticodon of an incoming aminoacyl tRNA base-pairs with the complementary mRNA codon in the A site.
2. The large ribosomal subunit attaches with the initiator tRNA with the amino acid methionine (Met) located in the P site.
1. The ribosome translocates the tRNA in the A site to the P site and the empty tRNA in the P site is moved to the E site where it is released.
3. A release factor, a protein shaped like an aminoacyl tRNA, promotes hydrolysis and releases the polypeptide.
Answer:
36 mol H
Explanation:
1 molecule C₆H₁₂O₆ contains 12 H atoms
1 mol C₆H₁₂O₆ contains 12 mol H atoms; Multiply by 3
3 mol C₆H₁₂O₆ contains 36 mol H atoms
If you are talking about moles of hydrogen molecules (H₂), you divide 36 by 2 and get 18 mol H₂.
Answer:
A redox reaction is spontaneous if the standard electrode potential for the redox reaction, Eo(redox reaction), is positive. ...
If Eo(redox reaction) is positive, the reaction will proceed in the forward direction (spontaneous)
Explanation:
thank me later
<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol
<u>Explanation:</u>
Rate of a gas is defined as the amount of gas displaced in a given amount of time.
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
So,
We are given:
Volume of unknown gas (X) = 1.0 L
Volume of oxygen gas = 1.0 L
Time taken by unknown gas (X) = 105 seconds
Time taken by oxygen gas = 31 seconds
Molar mass of oxygen gas = 32 g/mol
Molar mass of unknown gas (X) = ? g/mol
Putting values in above equation, we get:
Hence, the molar mass of unknown gas is 367.12 g/mol
Answer:
Let the volume of the block of wood be
V
c
m
3
and its density be
d
w
g
c
m
−
3
So the weight of the block
=
V
d
w
g
dyne, where g is the acceleration due to gravity
=
980
c
m
s
−
2
The block floats in liquid of density
0.8
g
c
m
−
3
with
1
4
t
h
of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid
=
1
4
V
×
0.8
×
g
dyne.
Hence by cindition of floatation
V
×
d
w
×
g
=
1
4
×
V
×
0.8
×
g
⇒
d
w
=
0.2
g
c
m
-3
,
Now let the density of oil be
d
o
g
c
m
-3
The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil =
60
%
×
V
×
d
o
×
g
dyne
Now applying the condition of floatation we get
60
%
×
V
×
d
o
×
g
=
V
×
d
w
×
g
⇒
60
100
×
V
×
d
o
×
g
=
V
×
0.2
×
g
⇒
d
o
=
0.2
×
10
6
=
1
3
=
0.33
g
c
m
−
3
Explanation:
