Answer:
CO = zero
CO2 =1 bar
O2 = 2.02 bar
Explanation:
We are given
initial pressure of CO = 1bar
total pressure = 3.52 bar
so initial pressure of O2 = 3.52 - 1 = 2.52 bar
the reaction is
2CO + O2 → 2CO2
using the unitary method
2 moles of CO2 → 1 mole of O2
1 bar of CO →
(required)
but we have more oxygen present , that means CO is the limiting reagent
- Final pressure of CO will be zero as it is the limiting reagent so it will be consumed completely
- 1 bar of CO →
of CO2 - 2.52 bar O2 (initially) - 0.5 bar (reacted) = 2.02bar O2
Answer : 0.026 moles of oxygen are in the lung
Explanation :
We can solve the given question using ideal gas law.
The equation is given below.

We have been given P = 21.1 kPa
Let us convert pressure from kPa to atm unit.
The conversion factor used here is 1 atm = 101.3 kPa.

V = 3.0 L
T = 295 K
R = 0.0821 L-atm/mol K
Let us rearrange the equation to solve for n.



0.026 moles of oxygen are in the lung
Explanation:
The atoms are chemically bonded together, and they retain their individual physical and chemical properties.
The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7
Answer:
5' RG GWCCY 3'
3' YCCWG GR 5'
Explanation:
The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).
The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'
target Sequence: 5' RGGWCCY 3'
3' YCCWGGR 5'
The enzyme cleavage point is at:
5' RG^GWCCY 3'
3' YCCWG^GR 5'
The product of the cleavage will give a sticky end Cleavage:
5' RG GWCCY 3'
3' YCCWG GR 5'
Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.