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Sav [38]
3 years ago
13

Stephans mother cuts a twig from a rose bush and plants it in the soil. After a few days, Stephan observes a new plant growing.

Which characteristic does the growth of the new plant depict?
Chemistry
2 answers:
mixer [17]3 years ago
8 0

Answer:

This depict that the growth of the new plant has taken place due to asexual form of reproduction.

Explanation:

A procedure in which new offspring is generated from a single parent without involving sex cells or gametes is known as asexual reproduction. One of the asexual forms of reproduction in plants is the artificial propagation of plants. The three general procedures of artificial propagation of plants are layering, cuttings, and grafting.  

In cutting, a novel plant is developed by incising a small section of a plant that can be anything comprising bud on it. This section of the plant is then watered and grown into soil, and after some days a new plant begins to grow.  

vazorg [7]3 years ago
6 0

Answer:

Asexual reproduction

Explanation:

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Calculate the density of nitrogen gas, in grams per liter, at stp.
lesantik [10]

Standard temperature is 273 K

Standard pressure is 1 atm

We use the ideal gas equation to find out density of nitrogen gas in g/L

Ideal gas equation:

PV = nRT\\  PV = (\frac{Mass}{Molar mass)}RT\\   P(Molar mass) = (\frac{Mass}{Volume})RT\\  \frac{Mass}{Volume}=\frac{P(molar mass)}{RT} \\  Density = \frac{P(Molar mass)}{RT}

Molar mass of N_{2} = 28 g/mol

Pressure = 1 atm

Temperature = 273 K

Density = \frac{(1atm)(28 g/mol)}{(0.08206 \frac{L.atm}{mol.K})(273 K)}

= 1.25 g/L

Therefore, density of nitrogen gas at STP is 1.25 g/L

4 0
3 years ago
Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m
mario62 [17]

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

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