When a runner participates in a marathon, they travel a distance of 26.2 miles. How many kilometers is

) PABA refers to para-aminobenzoic acid which is used in some sunscreen formulations. If a 0.055 M solution of PABA has a pH of

AleksandrR [38]

Answer: $K_a$ of PABA is 0.000022

Explanation:

$NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.055 M and $\alpha$ = ?

$K_a=?$

$pH=-log[H^+]$

$2.96=-log[H^+]$

$[H^+]=1.09\times 10^{-3}$

$[H^+]=c\times \alpha$

$1.09\times 10^{-3}=0.055\times \alpha$

$\alpha=0.02$

Putting in the values we get:

$K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}$

$K_a=0.000022$

Thus $K_a$ of PABA is 0.000022

5 0

3 years ago

Find the volume of 3.011x1023 molecules of O2 at STP.

lord [1]

Answer:

6.022 x 1023 molecules = 1 Mole O2 = 22.4L

3.022 x 1023 molecules = 1/2 Mole O2 = 11.2L

The answer is 11.2 Litres

Explanation:

8 0

2 years ago

Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fu

sasho [114]

Answer:

Oxidation by FAD

Explanation:

1. Oxidation by NAD⁺

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V

NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V

Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V

2. Oxidation by FAD

Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V

FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V

Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V

Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.

FAD is the stronger oxidizing agent.

The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.

8 0

3 years ago

Magnesium has a density of 1.738 g/cmº at 25°C. What is the mass of a block of

Rudiy27

Answer:

87.50g(1cm^3/1.1738g)=50.345 cm^3

Explanation:

volume units id in cubic centimeters

5 0

3 years ago

Calculate the molar solubility of ag2so4 in each solution below. the ksp of silver sulfate is 1.5 × 10−5. (a) 0.36 m agno3: (b)

Lana71 [14]

Answer:

a) 1.157 x 10⁻⁴ M.

b) 0.0032 M.

Explanation:

Ag₂SO₄ is sparingly soluble salt in water which is dissociate according to:

Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻.

Ksp = [Ag⁺][SO₄²⁻] = (2s)²(s) = 1.5 x 10⁻⁵.

4s³ = 1.5 x 10⁻⁵.

s³ = 1.5 x 10⁻⁵/4 = 3.75 x 10⁻⁶.

∴ s = ∛(3.75 x 10⁻⁶) = 0.0155 M.

a) 0.36 M AgNO₃:

The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of AgNO₃:

AgNO₃ → Ag⁺ + NO₃⁻.

So, the concentration of Ag⁺ will be that of AgNO₃ and Ag₂SO₄:

[Ag⁺] = (2s + 0.36)², s is neglected with respect to 0.36 M the concentration resulted from AgNO₃.

So, [Ag⁺] = (0.36)².

∴ Ksp = [Ag⁺][SO₄²⁻] = (0.36)²(s) = 1.5 x 10⁻⁵.

∴ s = 1.5 x 10⁻⁵/(0.36)² = 1.157 x 10⁻⁴ M.

b) 0.36 M Na₂SO₄:

The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of Na₂SO₄:

Na₂SO₄ → 2Na⁺ + SO₄²⁻.

So, the concentration of SO₄²⁻ will be that of Na₂SO₄ and Ag₂SO₄:

[SO₄²⁻] = (s + 0.36), s is neglected with respect to 0.36 M the concentration resulted from Na₂SO₄.

So, [SO₄²⁻] = (0.36).

∴ Ksp = [Ag⁺][SO₄²⁻] = (2s)²(0.36) = 1.5 x 10⁻⁵.

∴ 4s² = 1.5 x 10⁻⁵/(0.36) = 4.166 x 10⁻⁵.

∴ s² = 4.166 x 10⁻⁵/4 = 1.042 x 10⁻⁵.

∴ s = √(1.042 x 10⁻⁵) = 0.0032 M.

7 0

3 years ago