To find the number of electrons in a neutral gold atom look at it's atomic number. The atomic number is the number of protons in the element. For an atom to be neutral it needs the same number of electrons as protons, therefore the positive and negative charges will be balanced out. Gold's atomic number is 79 therefore it has 79 electrons.
Pressure decreases when you open the top. Pressure decreases, which favors the formation of H2CO3.
I better get your number.
The original concentration is 5.1 × 10⁹ CFU / ml
In order to attain a countable plate, the number of CFU must be present in between 10 and 200 per ml.
Let us take 0.1 ml and dilute it to 1 ml.
This minimizes the concentration to 5.1 × 10⁹ × 10⁻¹ = 5.1 × 10⁸ CFU/ml
In order to minimize the concentration in between 10 and 200, it can be reduced to 5.1 × 10¹
The final concentration = 5.1 × 10¹ CFU/ml
Initial concentration = 5.1 × 10⁹ CFU/ml
Volume of sample with 5.1 × 10¹ CFU = 5.1 × 10¹ CFU × (1 ml / 5.1 × 10⁹ CFU)
= 1.0 × 10⁻⁸ ml
This is the volume to be taken to obtain countable value, 51 CFU.
