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2 years ago

What are some non examples of biodiversity

2 answers:

8 0

A monocrop is a non example of biodiversity because it contains only one species, such as all corn, therefore there is very little biodiversity.

777dan777 [17]2 years ago

7 0

Here is the answer
There are multiple ways to look at the concept of biodiversity.
There is diversity on the genetic level within populations, diversity within ecosystems among the species that inhabit them, and even diversity among ecosystems.
You can also look at different scales of biodiversity. For example you can look at a square meter, and see 20 species in it and say that's a pretty diverse square meter. But then, if you look at the neighboring square meter and see the same 20 species, you might begin to question how diverse this ecosystem really is. In contrast, you can look at a square meter of another ecosystem and see 15 species, while its neighboring square meter has a completely different set of 15 species. Which is the ecosystem with greater biodiversity? It all depends on the scale you are interested in.

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The diagram shows the movement of particles from one end of the container to the opposite end of the container. A cylindrical co

trapecia [35]

Answer:

C. effusion because there is a movement of a gas through a small opening into a larger volume

Explanation:

Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.

This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.

3 0

2 years ago

Read 2 more answers

Convert 338 L at 63.0 atm to its new volume at standard pressure.

taurus [48]

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = $\frac{P1V1}{P2}$

Putting the values in the equation:

V2 = $\frac{338X63}{1}$

= 21294 L

as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

7 0

2 years ago

Which of the following is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid?

Leni [432]

Answer:

The correct answer is option A.

Explanation:

Volumetric flask : A glass ware with round lower body with flat bottom and with thin cylindrical neck along with mark which indicates the specific volume filled to that mark.It is used in preparation of standard solution of compound with desired concentration with fixed volume.

Erlenmeyer flask : is a flask with conical shape with flat bottom used in titration experiments to carry out reaction with fixed volume of solution.

Test tube : Small cylindrical tube with rounded bottom used to observe reaction in between reactant taken in small amount.

Graduated beaker : Laboratory glassware used measure larger volumes of solution or to mix or stir solutions and liquids.

Graduated cylinder : Laboratory thin cylindrical glassware with accurate marking of volume used to measure an accurate volume of solutions or liquids required in an experiment.

Volumetric flask is the best piece of laboratory glassware for preparing 500.0 mL of an aqueous solution of a solid

8 0

2 years ago

Read 2 more answers

If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years

Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually, radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

$\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g$

8 0

2 years ago

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

7 0

2 years ago