Answer:
pH = 5.47
Explanation:
The equilibrium that takes place is:
HIO ↔ H⁺ + IO⁻
Ka = ![\frac{[H+][IO-]}{[HIO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%2B%5D%5BIO-%5D%7D%7B%5BHIO%5D%7D)
= 2.3 * 10⁻¹¹
At equilibrium:
<u>Replacing those values in the equation for Ka and solving for x:</u>
Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47
Answer:
5.physical change
6.chemical change
7.physical change
8.conservation of mass
9.thermal energy
10.physical change
I honeslty dont know if this is right
explanation:
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.
Explanation:
speed = distance/time
= 23.7/54 m/s
= 0.44 m/s
speed of a dog running through a field = 0.44 m/s
After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
More about half-life: brainly.com/question/1160651
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