<span>Mass oxigen = 2.81 - 2.50 = 0.31 g
Moles O = 0.31 g / 15.9994 g/mol =0.0194
moles Cu = 2.50 g / 63.546 g/mol =0.0393
we divide by the smallest number
0.0194 / 0.0194 = 1 => O
0.0393 / 0.0194 = 2 => Cu
the formula is Cu2O ( Copper (I) oxide )
Ya got this</span>
Answer:
D has to be based on facts so d would be the answer and idea that can only be proven true those are facts not hypothesis or guesses
Explanation:
based on a body of facts that have been repeatedly confirmed through observation and experiment. Such fact-supported theories are not "guesses" but reliable accounts of the real world."
Answer: The concentration of 
is 0.08 M.
Explanation:-
According to the neutralization law,
where,
= molarity of 
solution = 0.105 M
= volume of solution = 25 ml
= molarity of solution = ?
= volume of solution = 31.5 ml
= valency of = 1
= valency of = 1
Therefore, the concentration of is 0.08 M.
Answer:
Explanation:
Hello!
In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:
Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:
Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:
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