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3 years ago

Is the following sentence true or false carbon atoms can only form a single bond between other carbon atomd?

1 answer:

7 0

Carbon atom has 4 valence electron and can form 4 bonds with carbon and other atoms and it can also form single, double and triple bond with other carbon atoms.
So, the statement is false.

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The maximum speed limit on a highway is 55 miles per hour. What is that speed in meters per second? (mi = 1.609 km)

Slav-nsk [51]

Answer:

24.5872 meters per second

Rouned 24.6 or 25

5 0

3 years ago

The element californium (cf) sells for $1000 per µg. assuming 6.02 x 1023 atoms of cf have a mass of 251 grams, how many atoms o

saveliy_v [14]

Answer:- $2.40*10^1^0atoms$

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

1 cent = 1 US penny

So, 1 US dollar = 100 US pennies

$1g=10^6\mu g$

Let's make the set up starting with 1 penny as:

$1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})$

= $2.40*10^1^0atoms$

Therefore, we can bye $2.40*10^1^0atoms$ of Cf in one US penny.

3 0

3 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

Is an example of the carbon cycle occurring slowly?(1 point)

o-na [289]

Decomposition is a part of the carbon cycle that occurs slowly hence movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter is a slow part of the carbon cycle.

<h3>What is the carbon cycle?</h3>

The carbon cycle is part of the biogeochemical cycles that exist in nature. It refers to the movement of carbon in the ecosystem. The carbon cycle cuts across the air, the land and the water bodies.

The process in the carbon cycle that occurs slowly among the options is the movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter.

Learn more about carbon cycle: brainly.com/question/1627609

7 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