The answer is 3.63. seconds.
Second order reaction is the reaction in which the rate of reaction depends on either the concentration of two reactant species or on the two times the concentration of single reactant species.
What is the integrated rate law for the second-order reaction?
- The integrated rate law that relates the concentration, time and rate constant for the second-order reaction is:
![\frac{1}{[A]} =\frac{1}{[A]_{0} } +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D%20%2Bkt)
Where
![\[\begin{array}{l}{\rm{[A] - concentration\ of\ reactant\ A\ at\ time\ t}}\\{{\rm{[A]}}_0}{\rm{ - initial\ concentration\ of\ reactant\ A}}\\{\rm{t - time}}\\{\rm{k - rate\ constant}}\end{array}\]](https://tex.z-dn.net/?f=%5C%5B%5Cbegin%7Barray%7D%7Bl%7D%7B%5Crm%7B%5BA%5D%20%20-%20%20concentration%5C%20of%5C%20reactant%5C%20A%5C%20at%5C%20time%5C%20t%7D%7D%5C%5C%7B%7B%5Crm%7B%5BA%5D%7D%7D_0%7D%7B%5Crm%7B%20-%20%20initial%5C%20concentration%5C%20of%5C%20reactant%5C%20A%7D%7D%5C%5C%7B%5Crm%7Bt%20-%20time%7D%7D%5C%5C%7B%5Crm%7Bk%20%20-%20%20rate%5C%20constant%7D%7D%5Cend%7Barray%7D%5C%5D)
- Now, in the given question,
k = 
![[NO_{2} ]= 0.62\ M](https://tex.z-dn.net/?f=%5BNO_%7B2%7D%20%5D%3D%200.62%5C%20M)
![[NO_{2} ]_{0} = 0.28\ M](https://tex.z-dn.net/?f=%5BNO_%7B2%7D%20%5D_%7B0%7D%20%3D%200.28%5C%20M)
- Thus, using the rate law, the time is calculated as-
Therefore,
- Hence, the it would take 3.63 seconds for the concentration of
to decrease from 0.62 M to 0.28 M if the reaction is second order.
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Answer:
6.8 mole of O₂
Explanation:
Given expression:
2H₂ + O₂ → 2H₂O
Number of moles of H₂ = 13.6moles
Unknown:
Number of moles of O₂ = ?
Solution:
In the given problem, we are to find the number of moles of oxygen gas that will use up 13.6mole of hydrogen gas;
From the reaction equation;
2 mole of H₂ will completely react with 1 mole of O₂
13.6 moles of H₂ will completely be used up by 
mole of O₂
= 6.8 mole of O₂
The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.
Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be 
g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.
Answer:c the correct technology cannot support this mission
Explanation:
