Chemistry !!!!! HELP !!!!!!!!
2 answers:
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Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
Hello!
The reaction between HBr and KOH is the following:
HBr+KOH→H₂O + KBr
To calculate the amount of HBr left after addition of KOH, you'll use the following equations:
![HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\ \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr](https://tex.z-dn.net/?f=HBr_f%3DHBr_i-KOH%3D%28%5BHBr%5D%2AvHBr%29-%28%5BKOH%5D%2AvKOH%29%20%5C%5C%20%20%5C%5C%20HBr_f%3D%280%2C25M%2A0%2C64L%29-%280%2C5M%2A0%2C32L%29%3D0%20mol%20HBr)
That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7.
Have a nice day!
The reaction between HBr and KOH is the following:
HBr+KOH→H₂O + KBr
To calculate the amount of HBr left after addition of KOH, you'll use the following equations:
That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be neutral, close to 7.
Have a nice day!