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3 years ago

How does the end point differ from the equivalence point of a titration?

Physics

1 answer:

Gwar [14]3 years ago

3 0

Answer:

Equivalence point and end point are terminologies in pH titrations and they are not the same. 

Explanation:

In a titration the substance added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the acid and base titrated determines the nature of the final solution.

At equivalence point the number of moles of the acid will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the pH at equivalence point. 

A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating neutral nature of the solution. 

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What is a hydraulic system

Simora [160]

Explanation:

Hydraulic systems use the pump to push hydraulic fluid through the system to create fluid power. The fluid passes through the valves and flows to the cylinder where the hydraulic energy converts back into mechanical energy. The valves help to direct the flow of the liquid and relieve pressure when needed

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3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

A transverse wave is traveling through a canal. If the distance between two successive crests is 3.17 m and four crests of the w

Thepotemich [5.8K]

Answer:

a. 0.18Hz

b. 0.56m/s

Explanation:

From the question we can deduct the following parameters

The wavelength, λ is define as the distance between two successful crest or trough and from the question we conclude that wavelength is 3.17m.

Also the period of the wave T can be computed as

T=22.6/4

T=5.65secs.

a. To compute the frequency, recall that frequency, F=1/period.

Hence,

F=1/5.65

F=0.18Hz

b. Next we compute the wave speed.

Wave speed=frequency *wavelength

Wave speed =0.18*3.17

Wave speed =0.56m/s

5 0

3 years ago

5) A parallel series electric blasting circuit consists of 6 branches of 8 caps each. Cap legwire length is 28 ft. The cap circu

andrezito [222]

Answer:

Minimum DC current $= 9$ Ampere

Voltage $= 74.862$ V

Explanation:

Here in this questions three resistors are used in series

Resistance of cap circuit

$R_1 =\frac{ 8* 2.3}{6} \\R_1 = 3.067$

Here $2.3$ is the resistance per cap

Resistance of bus wire

$R_2 = \frac{50 * 4.02 }{1000} \\R_2 = 0.201$

Where $4.02$ is the resistance per $1000$ feet of bus wire of gauge 16

Resistance of firing line

$R_3 = \frac{(2*1000)* 2.525}{1000} \\R_3 = 5.05$

where $2.525$ is the resistance per $1000$ feet of bus wire of gauge 14

Total resistance is equal to

$R = R_1 + R_2 + R_3\\R = 3.067 + 0.201 + 5.05 \\R = 8.318$

Current in firing line is equal to

$1.5 * 6\\= 9$

Voltage is equal to product of resistance and current

$= 9 * 8.318\\= 74.862$

4 0

2 years ago

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of

Alla [95]

Answer:

Change in momentum is 1.1275 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 274 g = 0.274 kg

It hits the floor and rebounds upwards.

The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s (-ve because the ball hits the ground)

It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)

We have to find the change in the ball's momentum. It is given by :

$\Delta p=p_f-p_i$

$\Delta p=m(v-u)$

$\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))$

$\Delta p=1.1275\ kg-m/s$

So, the change in the momentum is 1.1275 kg-m/s

3 0

2 years ago

How does the end point differ from the equivalence point of a titration?​

How does the end point differ from the equivalence point of a titration?