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2 years ago

What is the filtrate?

Chemistry

1 answer:

kondaur [170]2 years ago

5 0

Answer:

a liquid which has passed through a filter.

Explanation:

Filtration is a physical separation process that separates solid matter and fluid from a mixture using a filter medium that has a complex structure through which only the fluid can pass

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A 2.2 M solution is made by with 0.45 moles of a solute. What is the final volume of this solution?

Savatey [412]

Answer: The final volume of this solution is 0.204 L.

Explanation:

Given: Molarity of solution = 2.2 M

Moles of solute = 0.45 mol

Molarity is the number of moles of solute present divided by volume in liters.

$Molarity = \frac{no. of moles}{Volume (in L)}$

Substitute the values into above formula as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\2.2 M = \frac{0.45}{Volume}\\Volume = 0.204 L$

Thus, we can conclude that the final volume of this solution is 0.204 L.

7 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

PLEASE HURRY I'M TIMED

cricket20 [7]

You should give a picture of the fable!

5 0

2 years ago

Suppose these substance were placed in a freezer set to -50 grades celsius. Which substance would become a liquid. Clorine, Merc

Vadim26 [7]

Chlorine would become a liquid. Its boiling point is around -34 Celsius so at any temperature below that it would be liquid.

5 0

3 years ago

Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (

kherson [118]

Answer:

(a) sp³ sp³

H₃C - CH₃

(b) sp³ sp²

H₃C - CH = CH₂

sp²

H₃C - C ≡ C - CH₂OH

sp sp³

(d) sp³ sp²

H₃C - CH=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

sp³ sp³

H₃C - CH₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

sp³ sp²

H₃C - CH = CH₂

sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

sp³ sp

H₃C - C ≡ C - CH₂OH

sp sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.

Hence,

sp³ sp²

H₃C - CH=O

8 0

3 years ago