2*3.52*17
=<span>119.68 grams</span>
Neon is the element - Ne the symbol
Answer:
A. The sum of the voltages of the half-reactions is positive.
Explanation:
correct on a p e x
Answer:
3.1% is the fraction of the sample after 28650 years
Explanation:
The isotope decay follows the equation:
Ln[A] = -kt + Ln[A]₀
<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>
<em />
k could be obtained from Half-Life as follows:
K = Ln 2 / Half-life
K = ln 2 / 5730 years
K = 1.2097x10⁻⁴ years⁻¹
Replacing in isotope decay equation:
Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]
Ln[A] = -3.4657
[A] = 0.0313 =
<h3>3.1% is the fraction of the sample after 28650 years</h3>
<em />
Answer:
Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol
Explanation:
Time (t) = 4.00\;s
Initial concentration of NO2 = 1.33 M
Integrated law for second order reaction:
![\frac{1}{[A]}=\frac{1}{[A]_0} =kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%3Dkt)
Where, [A] = Concentration after time, t
[A]0 = Intitial concentration, k = rate constant, t = time
On substituting values in the above
![\frac{1}{[A]}=\frac{1}{1.33} =0.255 \times 4.00](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B1.33%7D%20%3D0.255%20%5Ctimes%204.00)
![\frac{1}{[A]} =1.772](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D1.772)
[A] = 0.5644 M
Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M
No. of mole = Molarity * volume
= 0.7656 * 1
= 0.7656 mol 0r 0.77 mol
