Explanation:
The given data is as follows.
Mass of antimony = 19.75 g
Molar mass of Sb = 121.76 g/mol
Therefore, calculate number of moles of Sb as follows.
Moles of Sb = 
= 
= 0.162 mol
Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.
Moles of oxygen =
= 
= 0.406 mol
Hence, ratio of moles of Sb and O will be as follows
Sb : O
1 : 2.5
We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.
Thus, we can conclude that the empirical formula of the given oxide is 
.
The molarity of KOH is 0.1055 M
<u><em> calculation</em></u>
Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH
H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O
step 2: find the moles of H₂C₂O₄.2H₂O
moles = mass÷ molar mass
from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol
= 0.2000 g ÷ 126 g/mol =0.00159 moles
step 3: use the mole ratio to calculate the moles of KOH
H₂C₂O₄.2H₂O : KOH is 1:2
therefore the moles of KOH =0.00159 x 2 = 0.00318 moles
step 4: find molarity of KOH
molarity = moles/volume in liters
volume in liters = 30.12/1000=0.03012 L
molarity is therefore = 0.00318/0.03012 =0.1055 M
Answer:
The volume of the vessel is 250 L
Partial pressure of hydrogen = 189 torr
Explanation:
Using Boyle's law
Given ,
V₁ = 20.0 L
V₂ = ?
P₁ = 25 atm
P₂ = 2 atm
Using above equation as:
<u>The volume of the vessel is 250 L.</u>
According to Dalton's law of partial pressure:-
So, according to definition of mole fraction:
Also,
Mole fraction of H₂ = 1 - Mole fraction of He = 1 - 0.75 = 0.25
So,
Total pressure = 756 torr
Thus,
<u>Partial pressure of hydrogen = 189 torr.</u>
The oxidation state of Mn2 is 2+
