Uh , what’s the question ?
Answer:

Explanation:
Whenever a question asks you, "What is the concentration after a given time?" or something like that, you must use the appropriate integrated rate law expression.
The reaction is 2nd order, because the units of k are L·mol⁻¹s⁻¹.
The integrated rate law for a second-order reaction is
![\dfrac{1}{\text{[A]}} =\dfrac{1}{\text{[A]}_{0}}+ kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%3D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt)
Data:
k = 2.4 × 10⁻²¹ L·mol⁻¹s⁻¹
[A]₀ = 0.0100 mol·L⁻¹
[A] = 0.009 00 mol·L⁻¹
Calculation
:
![\begin{array}{rcl}\dfrac{1}{\text{[A]}} & = & \dfrac{1}{\text{[A]}_{0}}+ kt\\\\\dfrac{1}{0.00900 }& = & \dfrac{1}{0.0100} + 2.4 \times 10^{-21} \, t\\\\111.1&=& 100.0 + 2.4 \times 10^{-21} \, t\\\\11.1& = & 2.4 \times 10^{-21} \, t\\t & = & \dfrac{11.1}{ 2.4 \times 10^{-21}}\\\\& = & \mathbf{4.6 \times 10^{21}}\textbf{ s}\\\end{array}\\\text{It will take $\large \boxed{\mathbf{4.6 \times 10^{21}}\textbf{ s}}$ for the HI to decompose}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B%5Ctext%7B%5BA%5D%7D_%7B0%7D%7D%2B%20kt%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B0.00900%20%7D%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B0.0100%7D%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C111.1%26%3D%26%20100.0%20%2B%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5C%5C%5C11.1%26%20%3D%20%26%202.4%20%5Ctimes%2010%5E%7B-21%7D%20%5C%2C%20t%5C%5Ct%20%26%20%3D%20%26%20%5Cdfrac%7B11.1%7D%7B%202.4%20%5Ctimes%2010%5E%7B-21%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BIt%20will%20take%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B4.6%20%5Ctimes%2010%5E%7B21%7D%7D%5Ctextbf%7B%20s%7D%7D%24%20for%20the%20HI%20to%20decompose%7D)
Answer:
It is true. a) 0.25 mol
Explanation:
<em>Hello </em><em>there?</em>
To begin solving this problem, you have to write down the chemical equation and make sure it is well balanced.
The chemical equation is;
3Mg(s) + N2(g) => Mg3N2(s)
1 mole of Mg = 24g
We have 18g of Magnesium (Mg) reacting with Nitrogen gas (N2)
From our equation,
Mole ratio = 3 : 1, (Mg : N2)
1 mol Mg = 24g
x mol Mg = 18g
x mol Mg = (18/24) = 0.75 mol Mg
But mole ratio = 3 : 1 (Mg : N2)
This means that 3 => 0.75 mol Mg
What about ratio 1 of N2?
N2 = (0.75 mol ÷ 3)/1
= 0.25 mol N2
<em>I </em><em>hope</em><em> </em><em>this </em><em>helps</em><em> </em><em>you </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Ha</em><em>v</em><em>e </em><em>a </em><em>nice </em><em>studies.</em><em> </em><em />
<span>R= 8.314 J/mol K
T= 273 + 102 = 375K</span><span>
= (3/2) x 8.314 x 375 = 4680 J/mol</span>