Answer:
a) 2NaOH+H2SO4→Na2SO4+2H2O2NaOH+H2SO4→Na2SO4+2H2O
b) Số phân tử NaOH : Số phân tử H2SO4 = 2:1
Số phân tử NaOH : Số phân tử Na2SO4 = 2:1
Số phân tử NaOH : Số phân tử H2O = 2:2
Explanation:
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
Answer:
a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
Explanation:
