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lorasvet [3.4K]
3 years ago
6

______ is the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Physics
1 answer:
Nuetrik [128]3 years ago
7 0

The ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

Explanation:

Percent ionization is used for quantifying the number of ions present in the weak acid when dissolved in a solution. So it is similar to the pKa value. The percent ionization value can be determined as negative log of dissociation constant. Also the as the number of ions increases in weak acid, the concentration of acid will be decreasing . It can be calculated using the formula for percent ionization as follows:

Percent ionization = \frac{Ionized acid concentration}{Initial concentration of acid} * 100

As the water volume or concentration increases, the acid will get diluted much more thus leading to decrease in the concentration of acid.

So the ratio of concentration of ionized acid to the initial concentration of acid multiplied by 100 will give the percent ionization of a weak acid in water increases as the concentration of acid decreases.

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A wire loop of radius 0.50 m lies so that an external magnetic field of magnitude 0.40 T is perpendicular to the loop. The field
vazorg [7]

The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

ℰ = emf, Δφ = change in magnetic flux, Δt = elapsed time

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φ = BA

B = magnetic field strength, A = loop area

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A = πr²

r = loop radius

Make a substitution:

φ = B2πr²

Since the strength of the magnetic field is changing while the radius of the loop isn't changing, the change in magnetic flux Δφ is given by:

Δφ = ΔB2πr²

ΔB = change in magnetic field strength

Make another substitution:

ℰ = |ΔB2πr²/Δt|

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3 0
3 years ago
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
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Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
3 years ago
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