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3 years ago

What happens to the atomic radius when an elctron is lost

Chemistry

1 answer:

Daniel [21]3 years ago

5 0

Answer:

It decreases.

Explanation:

Due to changes in the Coulombic force, the protons within the nucleus of the ion have a much easier time pulling at one fewer electrons. This way, the electrons are pulled closer to the center, causing the radius to decrease.

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Answer:

The concentration of $H^{+}$ is 1.48 × $10^{-9}$ M

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Explanation:

The pH of a solution is given by the formula below

pH = $-log_{10}[{H^{+}]$

∴ $[H^{+}] = 10^{-pH}$

where $[{H^{+}]$ is the $H^{+}$ concentration

From the question,

pH = 8.83±0.04

That is,

pH =8.83 and the uncertainty is ±0.04

First, we will determine $[{H^{+}]$ from

$[H^{+}] = 10^{-pH}$

$[{H^{+}] = 10^{-8.83}$

$[{H^{+}] = 1.4791$ × $10^{-9}$ M

$[{H^{+}] = 1.48$ × $10^{-9}$ M

The concentration of $H^{+}$ is 1.48 × $10^{-9}$ M

The uncertainty of $[{H^{+}]$ ( $U_{[H^{+}] }$ ) from the equation $[H^{+}] = 10^{-pH}$ is

$U_{[H^{+}] } = 2.303 \\$ × ${[H^{+}] }$ × $U_{pH }$

Where $U_{[H^{+}] }$ is the uncertainty of $[{H^{+}]$

$U_{pH }$ is the uncertainty of the pH

Hence,

$U_{[H^{+}] }$ = 2.303 × 1.4791 × $10^{-9}$ × 0.04

$U_{[H^{+}] }$ = 1.36 × $10^{-10}$ M

$U_{[H^{+}] }$ = 0.12 × $10^{-9}$ M

Hence, the absolute uncertainty of $[{H^{+}]$ is ±0.12 × $10^{-9}$ M

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3 years ago

What happens to the atomic radius when an elctron is lost​

What happens to the atomic radius when an elctron is lost