<span>When M(OH)2 dissolves we have
M(OH)2 which produces M2+ and 2OHâ’
pH + pOH=14
At ph =7; we have
7+pOH=14
pOH=14â’7 = 7
Then [OHâ’]=10^(â’pOH)
[OH-] = 10^(-7) = 1* 10^(-7)
At ph = 10. We have,
pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4)
Finally ph = 14. We have, pOH = 0
And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)...
So [OH-] = 1</span>
Answer:
29.3 g HClO₄
Explanation:
We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:
39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄
The answer is B
If one circuit fails, it is most likely that all the components in the circuit will fail.
The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams
The condensed formula would be CH3-CH(CH4)-CH2-CH(CH4)-CH2-CH(CH4)-CH3. The molecular formula would be C10H25.
