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earnstyle [38]
3 years ago
13

Determine the Ka value of propanoic acid if it is found that 2.95x10-3 M of each conjugate exists when a 0.65 M solution is made

.
Physics
1 answer:
MrRa [10]3 years ago
6 0

Answer:

1.34 × 10⁻⁵

Explanation:

Let's consider the acid dissociation of propanoic acid.

CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

We can find the acid dissociation constant (Ka) using an ICE chart.

      CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

I                 0.65                                           0                       0

C                  -x                                            +x                      +x

E               0.65 - x                                       x                        x

We know that [CH₃CH₂COO⁻] = [H₃O⁺] = x = 2.95 × 10⁻³ M

[CH₃CH₂COOH] = 0.65 - 2.95 × 10⁻³ = 0.64705 M

Ka = [CH₃CH₂COO⁻].[H₃O⁺]/[CH₃CH₂COOH]

Ka = (2.95 × 10⁻³)²/0.64705

Ka = 1.34 × 10⁻⁵

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2512.8 meters

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Which embedded molecule in the cell membrane moves large molecules into and out of the cell?
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The correct  answer is A.

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5 0
3 years ago
During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck
shutvik [7]
<h2>Answer:2.65 seconds</h2>

Explanation:

Let a be the acceleration.

Let u be the initial velocity.

Let v be the final velocity.

Let t be the time taken.

As we know from the equations of motion,

v=u+at

Given,v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}

0=8.12-3.06t

t=2.65sec

6 0
4 years ago
A marble statue has a mass of 6,200 grams and a volume of 2,296 cm3. What is the density of marble?
Elan Coil [88]
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6 0
3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0
3 years ago
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