The molecular weight of Mg(OH)2 : 58 g/mol
<h3>Further explanation</h3>
Given
Mg(OH)2 compound
Required
The molecular weight
Solution
Relative atomic mass (Ar) of element : the average atomic mass of its isotopes
Relative molecular weight (M) : The sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
So for Mg(OH)2 :
= Ar Mg + 2 x Ar O + 2 x Ar H
= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol
= 24 + 32 + 2
= 58 g/mol
Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm.
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
= 0.975 - (520/760)
= 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm
So the pressure of the gas =P(atm) + Height (Hg)
= 0.975 + (67/ 760) = 1.06 atm
Picture (3)
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
= (103/760) = 0.136 atm
What we are give: Concentration of base (CB) = 3.4 ×
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve.
CA=
