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3 years ago

Insoluble sulfide compounds are generally black in color.

Chemistry

1 answer:

AfilCa [17]3 years ago

5 0

<h3>Answer:</h3>

Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)

K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)

<h3>Explanation:</h3>

According to solubility rules, metal sulfides are insoluble except Calcium sulfide(CaS), Magnesium sulfide(MgS), Barium sulfide(BaS), and those of potassium, sodium, and ammonium.
Therefore, when all the other sulfides are formed during a reaction they form a precipitate that is shown to be in solid-state.
Precipitates are formed during precipitation reactions when cations and anions combine to form a compound that is insoluble in water.

In our case; In the equations given, some reactions will take place to form a precipitate while others will not occur.

That is;

Na₂S(aq)+KCl(aq) → No reaction
Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
Pb(ClO₃)₂(aq)+NaNO₃(aq) → No reaction
AgNO₃(aq)+KCl(aq) →AgCl(s) + KNO₃(aq)
K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)

The first reaction will not occur.

The second equation is a precipitation reaction that forms lead(ii)sulfide which is a black precipitate.

The third reaction will not take place.

The fourth reaction will be a precipitation reaction forming silver chloride precipitate.

The fifth equation is also a precipitation reaction that forms tin(ii) sulfide, SnS which is a black precipitate.

Therefore; the equations that answers our question are;

Li₂S(aq)+Pb(NO₃)₂(aq) → PbS(s) + LiNO₃(aq)
K2S(aq)+Sn(NO₃)₄(aq→ SnS(s) + KNO₃(aq)

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enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

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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

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