Hello there!
Hydrogen has one proton because the number of protons in an atom is equal to the atomic number of the element. S
ince hydrogen's atomic number is 1, it has one proton.
I hope that's help !
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
An intrinsic property is independent of how much of a material is present and is independent of the form of the material, one large piece or a collection of small particles. Intrinsic properties are dependent mainly on the fundamental chemical composition and structure of the material.
Answer:
13mL
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
HNO3 + KOH —> KNO3 + H2O
From the balanced equation above, we obtained the following data:
Mole ratio of the acid (nA) = 1
Mole ratio of the base (nB) = 1
Step 2:
Data obtained from the question.
This includes the following:
Molarity of the acid (Ma) = 6M
Volume of the acid (Va) =?
Volume of the base (Vb) = 39mL
Molarity of the base (Mb) = 2M
Step 3:
Determination of the volume of the acid.
Using the equation:
MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
6 x Va / 2 x 39 = 1/1
Cross multiply to express in linear form
6 x Va = 2 x 39
Divide both side by 6
Va = (2 x 39)/6
Va = 13mL
Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL
