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3 years ago

Is c2h2 polar or non polar?

Chemistry

2 answers:

Arturiano [62]3 years ago

6 0

The element c2h2 is polar.

Lyrx [107]3 years ago

3 0

Answer:

polar beacuse there is no way its non polar.

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PLEASE HELP ME SOLVE THIS.Thank you so much!

Tju [1.3M]

Answer: The coefficients for the given reaction species are 1, 6, 2, 3.

Explanation:

The given reaction equation is as follows.

$Cr_{2}O^{2-}_{7} + Cl^{-} \rightarrow Cr^{3+} + Cl_{2}$

Now, the two half-reactions can be written as follows.

Reduction half-reaction: $Cr_{2}O^{2-}_{7} + 3e^{-} \rightarrow Cr^{3+}$

This will be balanced as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$ ... (1)

Oxidation half-reaction: $Cl^{-} \rightarrow Cl_{2} + 1e^{-}$

This will be balanced as follows.

$6Cl^{-} \rightarrow 3Cl_{2} + 6e^{-}$ ... (2)

Adding both equation (1) and (2) we will get the resulting equation as follows.

$Cr_{2}O^{2-}_{7} + 14H^{+} + 6Cl^{-} \rightarrow 2Cr^{3+} + 3Cl_{2} + 7H_{2}O$

Thus, we can conclude that coefficients for the given reaction species are 1, 6, 2, 3.

6 0

3 years ago

What increases OH ions

hram777 [196]

Answer:

A base.

Explanation:

Basic solutions give OH- ions.

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 $[E_{total}]$