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3 years ago

what are simplified definitions of the following words"Synthesis, Decomposition, Single Replacement, or Double Replacement"?

Chemistry

1 answer:

faust18 [17]3 years ago

8 0

Answer:

The Four Major Types of Reactions

Name General Reaction Pattern

Combination or synthesis A + B ----> AB

Decomposition AB ----> A + B

Substitution or Single Replacement A + BC ----> B + AC

Metathesis or Double Displacement AB + CD ----> AD + CB

Combination or Synthesis Reactions Two or more reactants unite to form a single product.

S + O2 ---------> SO2

sulphur oxygen sulphur dioxide

2 S + 3 O2 ---------> 2 SO3

sulphur oxygen sulphur trioxide

2 Fe + O2 ---------> 2 FeO

iron oxygen iron (II) oxide

Decomposition Reactions A single reactant is decomposed or broken down into two or more

products.

CaCO3 ----------> CaO + CO2

calcium carbonate calcium oxide carbon dioxide

2 H2O -----------> 2 H2 + O2

water hydrogen oxygen

2 KClO3 -----------> 2 KCl + 3 O2

potassium chlorate potassium chloride oxygen

Substitution or Single Replacement Reactions A single free element replaces or is substituted for one of the elements in a compound. The free element is more reactive than the one its replaces.

Zn + 2 HCl ----------> H2 + ZnCl2

zinc hydrochloric acid hydrogen zinc chloride

Cu + 2 AgNO3 -----------> 2 Ag + Cu(NO3)2

copper silver nitrate silver copper (II) nitrate

H2 + 2 AgNO3 -----------> 2 Ag + 2 HNO3

hydrogen silver nitrate silver nitric acid

2 Na + 2 H2O -----------> 2 NaOH + H2

sodium water sodium hydroxide hydrogen

Metathesis or Double Displacement Reactions This reaction type can be viewed as an "exchange of partners." For ionic compounds, the positive ion in the first compound combines with the negative ion in the second compound, and the positive ion in the second compound combines with the negative ion in the first compound.

HCl + NaOH -----------> NaCl + HOH

hydrochloric sodium sodium water

acid hydroxide chloride

BaCl2 + 2 AgNO3 ----------> 2 AgCl + Ba(NO3)2

barium silver silver barium

chloride nitrate chloride nitrate

(precipitate)

CaCO3 + 2 HCl -----------> CaCl2 + H2CO3

calcium hydrochloric calcium carbonic

carbonate acid chloride acid

Explanation:

The vast number of chemical reactions can be classified in any number of ways. Under one scheme they can be categorized either as oxidation-reduction (electron transfer) reactions or non-oxidation-reduction reactions. Another completely different but common classification scheme recognizes

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3. If all the waves were measured for a period of 5 seconds, which wave had the highest frequency?

Vanyuwa [196]

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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