Answer:
Kw = 2.88 × 10⁻¹⁵
Explanation:
Let's consider the dissociation of water.
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
The equilibrium constant Kw is:
Kw = [H⁺].[OH⁻]
If pH = 7.27, we can find [H⁺]:
pH = -log [H⁺]
H⁺ = anti log (-pH) = anti log (-7.27) = 5.37 × 10⁻⁸ M
According to the balanced equation, 1 mole of H⁺ is produced per mole of OH⁻. So, [H⁺] = [OH⁻] = 5.37 × 10⁻⁸ M
Then,
Kw = [H⁺].[OH⁻]= (5.37 × 10⁻⁸)² = 2.88 × 10⁻¹⁵
A or B depends on what you mean by lit or glowing but when you place a wooden split in the sample the gas must reignite but there can be some confusion between hydrogen and oxygen mainly because a splint can cause a slight popping sound while it reignites but hydrogen pops are more violent and can most time extinguish the splint.
First, draw the 2-hexene. Th is is a molecule of six carbons with a double bond in the second carbon:
CH3 - CH = CH2 - CH2 - CH2 - CH3
Secong, put one Br on the second carbon and one Br on the third carbon:
CH3 - CBr = CBr - CH2 - CH2 - CH3
Third, cis means that the two Br are placed in opposed positions, this is drawn with one Br up and the other down. So, you need to represent the position of the Br in the space:
H Br H H H
| | | | |
H - C - C = C - C - C - C - H
| | | | |
H Br H H H
The important fact to realize is that the two Br are in opposed sides of the molecule.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
