The concentration of the sulfuric acid : 0.499 M
The net ionic equation
2
OH
⁻(aq]+
2
H
⁺
(aq]
→
2
H
₂O
(l]
<h3>Further explanation</h3>
Given
42.68 ml of 0.43 M KOH
18.40 ml of H2SO4
Required
the concentration of the sulfuric acid
the net ionic equation
Solution
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence
Input the value :
a=KOH, b= H2SO4
0.43 x 42.68 x 1 = Mb x 18.40 x 2
Mb = 0.499 M
The net ionic equation
Reaction
2KOH + H2SO4 → K2SO4 + 2H2O
2
K
⁺
(aq]+
2
OH
⁻
(aq]
+
2
H
⁺
(aq]
+
SO
₄²⁻(aq]
→
2
K
⁺
(aq]
+
SO
₄²⁻(aq]
+
2
H
₂
O
(l]
canceled the spectator ions :
2
OH
⁻(aq]+
2
H
⁺
(aq]
→
2
H
₂O
(l]
Maybe formulate a hypothesis....
Do you mean like what is the core for investigation?
Variable are factors that influence our observation for investigation.
Answer:
only japanese and english I understand sorry your languages
Explanation:
so sorry can you brainliest me pls I need Brainliest:)
Answer:
Explanation:
Its chemical formula is Fe2(SO4)3
Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
PV = nRT ; n = PV/RT
Substituting the known values,
n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
volume = 0.0833 L or 83.34 mL
