The delocalized cloud of π electrons in benzene is formed by the overlap of 6 ________ orbitals.

Chemistry

1 answer:

Radda [10]2 years ago

8 0

The delocalized cloud of π electrons in benzene is formed by the overlap of 6<u> p-orbitals</u>.

So, option B is correct one.

In the case of hydrocarbons, delocalisation occurs in benzene rings , where a hexagon of six carbon atoms has decalized electrons spread over the whole ring.

All of the carbon atoms in the benzene rings are $sp^{2}$ orbitals around the ring produces a framework of six sigma bonds, while the unhybridized p-orbitals which are perpendicular to this plane over in side-to-side fashion to form three pi-bonds.

To learn more about benzene

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Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0

4 years ago

A chemical reaction that gives off heat as it proceeds is said to be:.

Dafna1 [17]

Exothermic.

To remember this, ‘exo’ means outside, and ‘thermic’ means heat. It gives the outside (exo), heat (thermic).

7 0

3 years ago

5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points

labwork [276]

Answer: (A) $H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) $H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)$

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of $H_{3}O^{+}$ in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

$Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)$

$NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)$

Hence, net ionic equation will be as follows.

$H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)$

(B) When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

$NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)$