All categories

2 years ago

Which of the following covalent bonds is the strongest?

Chemistry

1 answer:

Basile [38]2 years ago

7 0

Answer:

a) quadruple covalent bond

Explanation:

Quadruple covalent bonds are the strongest covalent bonds. The more the number of bonding pairs, the stronger the bond becomes.

A covalent bond is formed by sharing of electrons between two species.
They are usually between compound with similar electronegativities.
The number of bonds between the atom determines the bond strength.

Single < double < triple < quadruple covalent bonds.

You might be interested in

quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe

12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

#SPJ4

6 0

1 year ago

After the solution reaches equilibrium, what concentration of zn2 (aq remains?

Fantom [35]

According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.

Thank you for your question. Please don't hesitate to ask in Brainly your queries.

8 0

2 years ago

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

2 years ago

molecules with a plane of symmetry between the chiral centers are achiral and meso. identify which of the following molecules ar

maks197457 [2]

Molecules with a plane of symmetry between the chiral centers are achiral and meso. From the given molecules (Picture attached) only (A) compound 1 is meso.

When compounds possess a plane of symmetry between the chiral centers they are called achiral or meso compounds. Among the given compounds (A) compound 1 have a plane of symmetry. So we can say compound one is a meso or achiral compound. Compounds two, three, and four have no plane of symmetry, as you can see in the structures attached. So all other compounds (compound 2, compound 3, and compound 4) except compound one are not meso or achiral.

You can also learn about meso compounds from the following question:

brainly.com/question/29022658

#SPJ4

7 0

10 months ago

Does changing the amplitude affect the period on a pendulum and how

yawa3891 [41]

Answer:

As the amplitude of pendulum motion increases, the period lengthens, because the restoring force −mgsinθ increases more slowly than −mgθ (sinθ≅θ−θ3/3!for small angles).

5 0

2 years ago