The reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Answer:
Covalent solids, also called network solids, are solids that are held together by covalent bonds. As such, they need localized electrons (shared between the atoms) and therefore the atoms are arranged in fixed geometries. Distortion far from this geometry can only occur through a breaking of covalent sigma bonds.
Answer:
H. 2 blue, 3 yellow, and 12 green
Explanation:
Aluminium atoms (Al) = Blue Beads
Oxygen Atoms (O) = Green Beads
Sulfur (S) = Yellow beads
From the compound Al2(SO4)3, the number of atoms present are;
Al = 2
S = 3
O = 12
This means the model would contain;
2 Blue beads
12 Green beads
3 Yellow beads
The correct option is; H. 2 blue, 3 yellow, and 12 green
Mg = 24.3
Cl = 35.5
24.3 + 35.5 x 2 = 95.3 ~ 95.21 ( all periodic tabes have different accuracies)
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Missing question: 0,535 gram of KIO₃ dissolved in 250 mL of de-ionized water to <span>make primary standard solution.
m(</span>KIO₃) = 0,535 g.
V(KIO₃) = 250 mL ÷ 1000 mL/L = 0,25 L.
n(KIO₃) = m(KIO₃) ÷ M(KIO₃).
n(KIO₃) = 0,535 g ÷ 214 g/mol.
n(KIO₃) = 0,0025 mol.
c(KIO₃) = n(KIO₃) ÷ V(KIO₃).
c(KIO₃) = 0,0025 mol ÷ 0,25 L.
c(KIO₃) = 0,01 mol/L = 0,01 M.
