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jeka94
2 years ago
5

Consider a culture medium on which only gram-positive organisms such as Staphylococcus aureus colonies can grow due to an elevat

ed NaCl level. A yellow halo surrounds the growth, indicating the bacterium fermented a sugar in the medium, decreasing the pH as a result and changing the color of a pH indicator chemical. This type
Chemistry
1 answer:
Olin [163]2 years ago
4 0
<h2>Selective & Differential Medium</h2>

Explanation:

  • Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
  • Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
  • Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria

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A cylinder container can hold 2.45 L of water. It’s radius is 4.00 cm. What is the volume of it in cubic centimeters?
tatuchka [14]

Answer:

2450 cm3

Explanation:

Volume of cylinder = V=πr2h

2.45L = 2450mL

1mL = 1 cm cubed

2450mL = 2450 cm cubed

7 0
2 years ago
We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find
iragen [17]

Explanation:

a)  Using the provided information about the density  of gold, the sample size, thickness, and the following  equations and comersion factors, find the area of   the gold leaf:

V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V

Gold _{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}

1 \mu=10^{-6} \mathrm{m}

First, find the volume of the sample and then find the area of the sample.

V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3} =\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}

V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}

b.  Using the provided information from part a ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}

h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}

8 0
2 years ago
If 8.0 g of NH4HS(s) is placed in a sealed vessel with a volume of 1.0 L and heated to 200 ∘C the reaction NH4HS(s) ⇌ NH3(g) + H
PIT_PIT [208]

Answer:

Increasing the volume of the vessel

Explanation:

By the Le Chatelier's principle, if a system in equilibrium suffer a variation that disturbs the equilibriu, the reaction shift in the way to minimize the pertubation and re-establish the equilibrium.

For a variation in pressure, when it increases, the reaction shift for the smallest of gas volume, and if decreases, the reaction will shift for the large gas volume. So, for the reaction given, the products have the large amount of gas, so by decreasing the pressure, more products will be formed, and the amount of NH₄HS will reduce. To decrease the pressure, we can increase the volume of the vessel: for the ideal gas equation (PV= nRT), pressure and volume are indirectly proportional.

7 0
1 year ago
A tank of oxygen has a volume of 40.0L and is held at a pressure of 159atm at 25∘C. What volume of O2 gas (in liters) would ther
Advocard [28]

Answer:

V2 = 6616 L

Explanation:

From the question;

Initial volume = 40L

Initial Pressure, P1 = 159atm

Initial Temperature T1 = 25 + 273 = 298K (Upon converting to Kelvin unit)

Final Volume, V2 = ?

Final Pressure, P2 = 1 atm

Final Temperature T2 = 37 + 273= 310K (Upon converting to Kelvin unit)

These quantities are related by the equation;

P1V1 / T1  = P2V2 / T2

V2 = T2 * P1 * V1 / T1 * P2

V2 = 310 * 159 * 40  /   (298 * 1)

V2 = 6616 L

6 0
1 year ago
Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
Stells [14]

Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0
2 years ago
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