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3 years ago

Which statement best describes the difference between a substance with a pH of 3.0 and a substance with a pH of 6.0?

1 answer:

7 0

The statement best describes the difference between a substance with a pH of 3.0 and a substance with a pH of 6.0 is - The substance with the lower pH has 1,000 times as many hydrogen ions per volume of water.

pH is the scale or measure for the substance about its acidic or basic strength. It ranges from 0 to 6.9 which is acidic and 7.1 to 14 which is basic.

Acidic substance has high concentration of Hydrogen ions whereas, basic substance has low concentration of hydrogen ions, however for the OH⁻ ions it is reverse.

Concentration of hydrogen ions is inversely related to its pH
More hydrogen ions present, the lower the pH
The fewer hydrogen ions, the higher the pH We know,

pH = -log( $H^{+}$ ) then,

=> 3 = -log ( $H^{+}$ )

=> $H^{+}$ = (for pH = 3)

=> pH = -log( $H^{+}$ )

=> 6 = -log ( $H^{+}$ )

=> $H^{+}$ = $10^{-6}$ (for pH = 6)

Thus, The substance with the lower pH (3) has 1000 times as many hydrogen ions per volume of water.

Learn more about pH scale:

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A sound wave travels through water. What best describes the direction of the water particles?

balu736 [363]

Answer:

In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next.

A water wave is an example of a transverse wave. As water particles move up and down, the water wave itself appears to move to the right or left.

7 0

3 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

Which are characteristics of electromagnetic waves?check all that apply.

emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, $c=3\cdot 10^8 m/s$ )

travel only through matter --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space --> FALSE, electromagnetic waves can also travel through matter

can bend around objects --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0

3 years ago

Read 2 more answers

Hi! I have a word bank I need help with please!<br> I have the questions as an attachment

Katena32 [7]

Scott needs to determine the density of a metallic rod. First, he should determine the mass of his sample on the laboratory balance. Second, he should measure the volume of his sample by water displacement. Finally, he can calculate the density by dividing mass/volume.
Hope this helped ;)

4 0

3 years ago