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3 years ago

As concentration increases by a tenth,what happens to the pH?

2 answers:

5 0

As the concentration of a solution, will increase by a tenth, the hydrogen ion concentration can become basic. The hydrogen ion concentration measures how acidic or basic the substance is, wherever zero is that the most acidic because of increase in H+ during a concentration, fourteen is that the most simple as a result of low H+ concentration and seven is taken into account neutral with an equal range of H+ and OH- ions.

Explanation:

pH may be a measure of the hydrogen particle concentration of a solution. Solutions with a high concentration of atomic number 1ions have an occasional hydrogen ion concentration and solutions with occasional concentrations of H+ ions have a high hydrogen ion concentration.

For a neutral solution, pH=7
Acidic solutions pH<7 
Basic solutions have pH>7

The lowers the pH, a lot of acidic the solution and therefore the higher the concentration of H+.

charle [14.2K]3 years ago

4 0

As concentration of a solution, increases by a tenth, the pH will become basic. The pH scale measures how acidic or basic the substance is, where 0 is the most acidic due to increase in H+ in a concentration, 14 is the most basic because of low H+ concentration and 7 is considered neutral with equal number of H+ and OH- ions.

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How are density and pressure related related to each other?

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Answer:

The pressure is the measure of force acting on a unit area. Density is the measure of how closely any given entity is packed, or it is the ratio of the mass of the entity to its volume. The relation between pressure and density is direct. Change in pressure will be reflected in a change in density

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3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

2 years ago

You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment

Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0

2 years ago

Which of the following diagrams correctly shows the electron configuration of Sulfur, with atomic number 16?

nika2105 [10]

Answer:

Diagram C

Explanation:

We are given that Sulfur with atomic number 16.

We have to find that which diagram shows the electronic configuration of sulfur.

S=16

Its Diagram C

6 0

3 years ago

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