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vredina [299]
2 years ago
11

127) Thirty-six colonies grew in nutrient agar from 1.0 ml of undiluted sample in a standard plate count. How many cells were in

the original sample?
Chemistry
1 answer:
SIZIF [17.4K]2 years ago
6 0

Answer:

36

Explanation:

Since the sample was undiluted the number of colonies is the number that grew on the nutrient agar which is 36 colonies. If it was diluted for example let say 0.1 ml from a dilution in which 1 ml of the sample was added to 9 ml of water, and it grew  colonies then  0.1 ml  yielded  6 colonies, 1 ml of the diluted sample will yield 60 colonies and 10 ml will have 600 colonies and therefore the 1 ml undiluted sample will have 600 colonies.

You might be interested in
What is the concentration of Iodine I2 molecules in a solution containing 2.54 g of iodine 250 cm3 of solution? A 0.01mol/dm3 B
gulaghasi [49]

Answer:

C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine (c), measured in moles per cubic decimeter, can be determined by the following formula:

c = \frac{m}{M\cdot V} (1)

Where:

m - Mass of iodine, measured in grams.

M - Molar mass of iodine, measured in grams per mol.

V - Volume of solution, measured in cubic decimeters.

If we know that m = 2.54\,g, M = 253.809\,\frac{g}{mol} and V = 0.25\,dm^{3}, then the concentration of iodine in a solution is:

c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}

c = 0.04\,\frac{mol}{dm^{3}}

Hence, the correct answer is C.

3 0
2 years ago
Express each aqueous concentration in the unit indicated.
MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L =  9500 millileters (ml)

we consider 25 mg = 25 millileters

<em>ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)</em>

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml =  10000  × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

<em>ppm = (mass of solute / mass of solvent) × 1000000 (1 million)</em>

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

<em>moles = mass / molar mass</em>

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

<em>Molarity = moles of solute / litres of solution</em>

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0
2 years ago
In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are
mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

W = 200*(1-0.704)\\W=59.2\ g

There are 59.2 grams of water in this solution.

5 0
2 years ago
Why carbon dioxide in the atmosphere appears to be increasing
Mumz [18]

Answer:

Explanation:

Burning fossil fuels releases the carbon dioxide stored millions of years ago. ... The concentration of carbon dioxide in the atmosphere has increased more in the northern hemisphere where more fossil fuel burning occurs. Since the Industrial Revolution the concentration globally has increased by about 40 % .

4 0
2 years ago
Read 2 more answers
Bismuth(III) sulfate is an ionic compound formed from Bi3+ and SO42-. What is the correct way to represent the formula?
Hitman42 [59]

Answer:

Bi2(SO4)3

Explanation:

Bismuth(iii) sulfate is an ionic compound therefore, their is transfer of electron. Ionic compound has both cations and anions. The cations is positively charged ion while the anions is negatively charged ions. The cations loses electron to become positively charged while the anions gains electron to become negatively charged.

From the compound above, Bismuth(iii) sulfate the cations will be Bismuth ion which loses 3 electrons. The anions is the sulfate ion (S04)2- with a -2 charge.

The chemical formula can be computed from the charge configuration as follows

Bi3+  and (SO4)2-

cross multiply the charges living the sign behind to get the chemical formula

Bi2(SO4)3

Note the final chemical formula, the numbers are sub scripted

4 0
2 years ago
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