Answer:
The oxygen will appear in the final stage if the oxygen is excess from the beginning.
Explanation:
1. C +(1/2)O2 —> CO
2. CO+(1/2)O2—> CO2
The first stage explains that the oxygen is limited and hence CO is produced.
The 2nd stage explains that if the oxygen is excess, then CO2 will be produced.
The overall reaction when oxygen is excess is given by:
C + O2 —> CO2
Analogous structures are those structures in different species which perform the same function, have similar appearance and structure but are not evolved together; therefore do not share a common ancestor. Homologous and analogous organs video explains in a simple way.
Answer:
2. Option B.
Explanation:
H₂SO₄ + Ba(OH)₂ → BaSO₄ + 2H₂O
You can count 2H in sulfuric acid and 2 H in the barium hyrdoxide, so the coefficient for water must be 2.
You will have 4 H on both sides of the reaction.
Try with the dissociations of each reactant
Sulfuric acid ⇒ H₂SO₄ → 2H⁺ + SO₄⁻²
Barium hydroxide ⇒ Ba(OH)₂ → Ba²⁺ + 2OH⁻
Sulfate anion bonds to barium cation to produce the salt, therefore the 2 protons will bond the 2 hydroxide in order to produce, 2 moles of H₂O
2H⁺ + 2OH⁻ → 2H₂O
Answer:
0.071 is the correct answer using the formula d=m/v
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
