How many total moles of ions are released when each of the following dissolves in water?

Chemistry

1 answer:

Gnom [1K]2 years ago

5 0

Thus 3 moles of ions are released when samples dissolves definitely in water.

Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen. A mole of water molecules would be 2 moles of hydrogen atoms plus 1 mole of oxygen atoms.

One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The variety 6.022 × 10²³ is acknowledged as Avogadro's number or Avogadro's constant.

<h3>How plenty is a mole of water?</h3>

18.0146 grams

The atomic mass of hydrogen is 1.0078 and the atomic mass of oxygen is 15.999, so the mass of a mole of water is two x 1.0078 + 15.999. Therefore, a mole of water, or 6.022 x 1023 molecules, weighs 18.0146 grams.

Learn more about moles of ion here:

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Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

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(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$