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3 years ago

Which are examples of habitat destruction? Check all that apply

Chemistry

2 answers:

Rom4ik [11]3 years ago

6 0

Answer:

Deforestation

Pollution

Poaching

Explanation:

RSB [31]3 years ago

5 0

An example of habitat destruction could be a squirrel who lives in a tree but modern civilication cutting down the squirrels tree and leaving it homeless

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What is the speed of light?can a object move in speed of light?

timurjin [86]

Answer:

The speed of light is the speed at which light travels. No, an object cannot move at the speed of light.

Explanation:

The speed of light is 186,000 miles per second. An object with mass cannot move at the speed of light since it would take an infinite amount of energy to achieve that velocity, since only massless particles can travel at the speed of light. Also, you would have to factor in air friction, meaning even if an object were to reach such high speeds, it would instantly disintegrate.

4 0

2 years ago

What is the mole fraction of O2O2 in a mixture of 15.1 gg of O2O2, 8.19 gg of N2N2, and 2.46 gg of H2H2

SVETLANKA909090 [29]

Answer:

Mole fraction O₂= 0.43

Explanation:

Mole fraction is the moles of gas/ total moles.

Let's determine the moles of each:

Moles O₂ → 15.1 g / 16 g/mol = 0.94

Moles N₂ → 8.19 g / 14 g/mol = 0.013

Moles H₂ → 2.46 / 2 g/mol = 1.23

Total moles = 2.183

Mole fraction O₂= 0.94 / 2.183 → 0.43

3 0

3 years ago

Grams of cl in 38g of cf3cl

Lerok [7]

Answer:

114 grams

Explanation:

3chlorines per compound*38grams=114

8 0

3 years ago

What is the density of a plastic ball that has a volume of 6cm) and a mass of 12g?<br> 50 cm3

marissa [1.9K]

Answer:

2 g/cm³

Explanation:

4 0

2 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