Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.

<u>Answer:</u> The concentration of $SO_3$ in the vessel after 0.240 seconds is 0.24 M

<u>Explanation:</u>

For the given chemical equation:

$2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)$

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $14.1M^{-1}s^{-1}$

t = time taken= 0.240 second

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 1.44 M

Putting values in above equation, we get:

$14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)$

$[A]=0.245M$

Hence, the concentration of $SO_3$ in the vessel after 0.240 seconds is 0.24 M

4 0

3 years ago

Nitrogen has two isotopes. One has an atomic mass of 14.003 amu and a relative abundance of 99.63% while the other isotope has a

Slav-nsk [51]

Answer:

Average atomic mass = 14.0067 amu.

Explanation:

Given data:

Abundance of 1st isotope = 99.63%

Atomic mass of 1st isotope = 14.003 amu

Abundance of 2nd isotope = 0.37%

Atomic mass of 2nd isotope = 15.000 amu

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (99.63×14.003)+(0.37×15.000) /100

Average atomic mass = 1395.119 + 5.55 / 100

Average atomic mass = 1400.67 / 100

Average atomic mass = 14.0067 amu.

3 0

3 years ago