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2 years ago

Which plate has the most countable

Chemistry

1 answer:

Rama09 [41]2 years ago

3 0

The agar plate with the most countable colonies should be determined by using a colony forming unit (CFU) calculator.

<h3>What is a colony?</h3>

A colony can be defined as a group of visible mass of microorganisms such as bacteria, fungi, that originated (cultured) from a single-mother cell and are grown on a solid agar medium.

In order to determine the most countable colonies, we have to count the total number of colonies by using a hemacytometer and then use a colony forming unit (CFU) calculator to determine the cfu/mL.

<u>Note:</u> You should choose a dilution factor based on the total number of colonies contained in the agar plate.

Finally, input the correct dilution factor into the colony forming unit (CFU) calculator to determine which agar plate has the most countable colonies.

Read more on dilution factor here: brainly.com/question/24881505

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3 years ago

Given that the wavelength of a wave is 4.57 x 10^9 m, what must be the

r-ruslan [8.4K]

Answer:

Option d is correct option = Frequency of light is 6.56×10⁻² Hz

Explanation:

Given data:

Frequency of light = ?

wavelength of light = 4.57×10⁹m

Solution:

Formula:

Speed of light = wavelength × frequency

c = λ × f

f = c/λ

This formula shows that both are inversely related to each other.

The speed of light is 3×10⁸ m/s

Frequency is taken in Hz.

It is the number of oscillations, wave of light make in one second.

Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.

Now we will put the values in formula.

f = 3×10⁸ m/s / 4.57×10⁹m

f = 0.656×10⁻¹s⁻¹

s⁻¹ = Hz

f = 0.656×10⁻¹ Hz or 6.56×10⁻² Hz

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4 years ago

Banana dipped in liquid experiments

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4 years ago

Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide: PbCO 3 (s) PbO (s) CO 2 (g) ________ grams of lead (I

blsea [12.9K]

Answer:

We will have 7.30 grams lead(II) oxide

Explanation:

Step 1: Data given

Mass of lead (II)carbonate = 8.75 grams

Molar mass PbCO3 = 267.21 g/mol

Step 2: The balanced equation

PbCO3 (s) ⇆ PbO(s) + CO2(g)

Step 3: Calculate moles PbCO3

Moles PbCO3 = mass / molar mass

Moles PbCO3 = 8.75 grams / 267.21 g/mol

Moles PbCO3 = 0.0327 moles

Step 4: Calculate moles PbO

For 1 mol PbCO3 we'll have 1 mol PbO and 1 mol CO2

For 0.0327 moles PbCO3 we'll have 0.0327 moles PbO

Step 5: Calculate mass PbO

Mass PbO = moles PbO * molar mass PbO

Mass PbO = 0.0327 moles * 223.2 g/mol

Mass PbO = 7.30 grams

We will have 7.30 grams lead(II) oxide

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4 years ago