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3 years ago

Which atom, selenium (Se) or Arsenic (As), has the higher electron affinity and why?

Chemistry

1 answer:

IrinaVladis [17]3 years ago

8 0

Answer:

5. Selenium, because it does not have a stable, half-filled p subshell and adding an electron does not decrease its stability.

Explanation:

Electron affinity is the amount of energy released when an isolated gaseous atom accepts electron to form the corresponding anion.

Selenium:-

The electronic configuration of the element is:-

$[Ar]3d^{10}4s^24p^4$

Arsenic:-

The electronic configuration of the element is:-

$[Ar]3d^{10}4s^24p^3$

The 4p orbital in case of arsenic is half filled which makes the element having more stability as compared to selenium.

Thus, selenium has higher electron affinity because adding electron does not decrease the stability as in case of arsenic.

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What is the molarity of a solution made by mixing 75.0 mL of 3.00 M NH₄OH with enough water to give 250. mL of solution?

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Mass 1=3M

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By using molarity formula:

<u>mass1*volume 1=mass2*volume 2</u>

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Describe how to prepare 400 ml of .85 m hcl using concentrated 17 m stock hcl

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You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.2750 M NaOH sol

oksano4ka [1.4K]

Answer:

a. 0.393M CH₃COOH.

b. 2.360% of acetic acid in the solution

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

<em>That means 1 mole of acid reacts per mole of NaOH.</em>

Moles of NaOH to reach the equivalence point are:

35.75mL = 0.03575L × (0.2750mol / L) = <em>9.831x10⁻³ moles of NaOH</em>

As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.

a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:

9.831x10⁻³ moles / 0.02500L =

b. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:

9.831x10⁻³ moles ₓ (60g / mol) = <em>0.590g of CH₃COOH</em>.

As volume of the solution is 25.00mL, the percentage of acetic acid is:

(0.590g CH₃COOH / 25.00mL) ₓ 100 =

<h3>2.360% of acetic acid in the solution</h3>

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3 years ago