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3 years ago

What are four main types of fresh water ecosystems

Chemistry

2 answers:

saw5 [17]3 years ago

8 0

Answer:

streams and rivers

ponds and lakes

wetlands

ocean gulf of Mexico

hope it helped

tensa zangetsu [6.8K]3 years ago

4 0

Oceans. Oceans are some of the most diverse and geographically expansive ecosystems on earth. Ocean ecosystems are divided into four zones: intertidal, pelagic, benthic and abyssal. The intertidal zone comprises the regions where ocean waters meet land.

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Given the formula: what is a chemical name of this compound? propane propanal propanol propanone

katrin [286]

1) Chemical formula for propane is CH₃-CH₂-CH₃.
Propane is a three carbon alkane (acyclic saturated hydrocarbon).
2) Chemical formula for propanal is CH₃-CH₂-CH=O.
Propanal is a saturated three carbon aldehyde (have a carbonyl center).
3) Chemical formula for propanol is CH₃-CH₂-CH₂-OH.
1-propanol is a primary alcohol.
4) Chemical formula for propanone is (CH₃)₂-C=O.
Propanone or acetone is he simplest and smallest ketone.

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3 years ago

The starting position of this object is m. The object is traveling at a velocity of m/s.

Ad libitum [116K]

Answer:

The starting position of this object is 3 m

The object is traveling at a velocity of 3 m/s

Explanation:

6 0

3 years ago

Read 2 more answers

Explica brevemente cuidado del aparato digestivo

kirill115 [55]

Answer:

For better digestive health, follow these simple tips:

Eat a high-fiber diet. ...

Be sure you're getting both soluble and insoluble fiber. ...

Minimize your intake of foods high in fat. ...

Select lean meats. ...

Add probiotics to your diet. ...

Follow a regular eating schedule. ...

Drink plenty of water.

Explanation:

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3 years ago

How many sig figs 18.01528

Ulleksa [173]

Answer:7

Explanation:

5 0

3 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago