Answer:
A cation is formed when a metal ion loses a valence electron while an anion is formed when a non-metal gains a valence electron. They both achieve a more stable electronic configuration through this exchange.
Just post the question on here
Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,
<span>The reason it will be 7 for some titrations is that when you titrates a strong acid with a strong base for example HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7
second reason is
When you titrates a weak acid with strong base at equivalence point
only a water solution of the conjugate base exists
CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
