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Arte-miy333 [17]
3 years ago
5

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

^ + ions are there in a solution at a pH = 2 than in a solution at a pH = 6?
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

The concentration of hydrogen ion at pH is equal to 2 := [H^+]=0.01 mol/L

The concentration of hydrogen ion at pH is equal to 6 : [H^+]'=0.000001 mol/L

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

pH=-\log [H^+]

The hydrogen ion concentration at pH is equal to 2 = [H^+]

2=-\log [H^+]\\

[H^+]=10^{-2}M= 0.01 M=0.01 mol/L

The hydrogen ion concentration at pH is equal to 6 = [H^+]

6=-\log [H^+]\\\\

[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L

Concentration of hydrogen ion at pH is equal to 2 =[H^+]=0.01 mol/L

Concentration of hydrogen ion at pH is equal to 6 = [H^+]'=0.000001 mol/L

The difference between hydrogen ion concentration at pH 2 and pH 6 :

= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L

Moles of hydrogen ion in 0.009999 mol/L solution :

=0.009999 mol/L\times 1 L=0.009999 mol

There are 0.009999 more moles of  H^+ ions in a solution at a pH = 2 than in a solution at a pH = 6.

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At a pressure of one atmosphere methane boils at −161°C and freezes at −182.5°C. Consider a temperature scale where the boiling
lawyer [7]

Answer:

-421.627906977\^{\circ}M

Explanation:

Absolute zero in the Kelvin scale = -273.15 °C

Temperature difference in the boiling and freezing point

-161-(-182.5)=21.5\ ^{\circ}C

Change in 21.5°C = Change of 100°C in M

1^{\circ}C=\dfrac{100}{21.5}=4.6511627907\ ^{\circ}M

Difference in the Celsius scale

-182.5-(-273.15)=90.65\ ^{\circ}C

Change in 90.65\ ^{\circ}C corresponds to

4.6511627907\times 90.65=421.627906977\ ^{\circ}M

Therefore the absolute zero in the Methane scale is

-421.627906977\ ^{\circ}M

3 0
3 years ago
A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i
geniusboy [140]

Answers:

a) t=0.311 s

b) a=86.847 m/s^{2}

c) y=1.736 m

d) V=17.369 m/s

Explanation:

For this situation we will use the following equations:

y=y_{o}+V_{o}t+\frac{1}{2}at^{2} (1)  

V=V_{o} + at (2)  

Where:  

y is the <u>height of the model rocket at a given time</u>

y_{o}=0 is the i<u>nitial height </u>of the model rocket

V_{o}=0 is the<u> initial velocity</u> of the model rocket since it started from rest

V is the <u>velocity of the rocket at a given height and time</u>

t is the <u>time</u> it takes to the model rocket to reach a certain height

a is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

V=\frac{V_{1}+V_{2}}{2} (3)

For this rocket is:

V=\frac{27 m/s}{2}=13.5 m/s (4)

Time is determined by:

t=\frac{y}{V} (5)

t=\frac{4.2 m}{13.5 m/s} (6)

Hence:

t=0.311 s (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

y=\frac{1}{2}at^{2} (8)  

We will use y=4.2 m :

4.2 m=\frac{1}{2}a(0.311)^{2} (9)  

Finding a:

a=86.847 m/s^{2} (10)  

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again y=\frac{1}{2}at^{2} but for t=0.2 s:

y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2} (11)

y=1.736 m (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

V= at (13)  

V= (86.847 m/s^{2})(0.2 s) (14)  

Finally:

V=17.369 m/s (15)  

5 0
3 years ago
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you
Ilya [14]

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W

So, the required power is 11666.6 W.

4 0
3 years ago
An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

  • a) \vec{v} = 70·i
  • b) The package will reach ground in approximately <u>12.77 seconds</u>.
  • c) The speed of the package as it lands is approximately <u>145.51 m/s</u>.
  • d) The path of the package based on a stationary frame of reference is <u>parabolic</u>
  • e) The path of the package as seen from the plane is <u>directly vertical</u> downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, \vec{v} = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}

Which gives;

\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, v_y, is given as follows;

v_y^2 = 2·g·h

Therefore;

v_y = √(2·g·h)

Which gives;

v_y = √(2 × 9.81 × 800) ≈ 125.28

v_y ≈ 125.28 m/s

Which gives; \vec{v} = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ <u>143.51 m/s</u>

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

6 0
2 years ago
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
Tasya [4]

Answer:

C) T

Explanation:

M = Mass attached to an ideal spring

A = Amplitude of the motion

T = Time period of oscillation

k = Spring constant of the spring

A = Amplitude of the motion

Time period of oscillation of the mass attached to the spring is given as

T = 2\pi \sqrt{\frac{M}{k} }

So we see that the time period does not depend on the amplitude. hence the period of oscillation remains the same.

8 0
4 years ago
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