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3 years ago

A sample of carbon 12 has a mass. Of 6.00g how many atoms of carbon 12 are in the sample

1 answer:

7 0

The number of atoms = number of moles x Avogadro’s number (6.022x10^23)

So first you need to find the number of moles of carbon 12

Number of moles = mass (g)/ molar mass

n= 6/12
= 0.5

Then put that into the equation for number of atoms so
Number of atoms = 0.5 x 6.022 x 10^23
= 3.011x10^23

I’m not 100% sure if that’s correct though

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Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is

Elis [28]

Answer:

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products =

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

We have :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

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

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

8 0

3 years ago

Why cant you see individual atoms without a special tool

lesya692 [45]

Answer:

its because atoms are incredibly small its looking for atoms is like placing a blueberry in a foot ball field and looking at it from 10 miles up you cant see that blueberry

Explanation:

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earnstyle [38]

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4 years ago

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irakobra [83]

V2 = 250 ml

Explanation:

Given:

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3 years ago

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The answer is Conduction

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