Explanation:
The reaction equation will be as follows.

Calculate the amount of
dissolved as follows.

It is given that
= 0.032 M/atm and
=
atm.
Hence,
will be calculated as follows.
=
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
= 
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
Cat urine. May sound weird, may get it wrong, but it's true. Your teacher'll freak out.
Answer:
Blue
Explanation:
If you look at a flame, blue is always at the bottom right? So that would be common sense that blue would be the hottest.
3.124mg of I-131 is present after 32.4 days.
The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.
What is Half life?
The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.
Half of the iodine-131 will still be present after 8.1 days.
The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.
The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.
If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.
Learn more about the Half life of radioactie element with the help of the given link:
brainly.com/question/27891343
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