Answer:
2

Explanation:
Half-life


Concentration
![{[A]_0}_A=1.2\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_A%3D1.2%5C%20%5Ctext%7BM%7D)
![{[A]_0}_B=0.6\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_B%3D0.6%5C%20%5Ctext%7BM%7D)
We have the relation
![t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%5Cpropto%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%5E%7Bn-1%7D%7D)
So
![\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%7Bt_%7B1%2F2%7D%7D_A%7D%7B%7Bt_%7B1%2F2%7D%7D_B%7D%3D%5Cleft%28%5Cdfrac%7B%7B%5BA%5D_0%7D_B%7D%7B%7B%5BA%5D_0%7D_A%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B2%7D%7B4%7D%3D%5Cleft%28%5Cdfrac%7B0.6%7D%7B1.2%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7Bn-1%7D)
Comparing the exponents we get

The order of the reaction is 2.
![t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7Bt_%7B1%2F2%7D%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7B2%5Ctimes%201.2%5E%7B2-1%7D%7D%5C%5C%5CRightarrow%20k%3D0.4167%5C%20%5Ctext%7BM%7D%5E%7B-1%7D%5Ctext%7Bmin%7D%5E%7B-1%7D)
The rate constant is 
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:
are a gas at very low volumes, when gas particles are very close together
a gas at very low temperatures, when gas particles have very little kinetic energy
a gas with highly polar molecules that have very strong intermolecular forces
Explanation:
Answer:
D
Explanation:
If the pressure remains constant then the temperature and Volume are all that you have to consider.
Givens
T1 = 19oC = 19 + 273 = 292o K
T2 = 60oC = 60 + 273 = 333oK
V1 = 250 mL
V2 = x
Formula
V1/T1 = V2/T2
250/292 = x/333
Solution.
The solves rather neatly. Multiplly both sides by 333
250*333 / 292 = 333 *x / 333
Do the multiplication
250 * 333 / 292 = x
83250 / 292 = x
Divide by 292
x = 285.1 mL
The answer is D
The answer is strong winds