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gulaghasi [49]
2 years ago
6

In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed

in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C. What is the concentration of H+ ions at a pH = 2? mol/L What is the concentration of OH– ions at a pH = 2? mol/L What is the ratio of H+ ions to What is the concentration of H+ ions at a pH = 11?
Physics
2 answers:
balu736 [363]2 years ago
5 0

The Answers are:

1) 0.01 mol/L

2) 0.000000000001 mol/L

3) 10,000,000,000 : 1

Liono4ka [1.6K]2 years ago
3 0
You must know the concept of pH of a solution and its relation to the concentration of H+ and OH- ions. pH is a measure of the substance's acidity or basicity. From the definition of Arrhenius, an acid contains an H+ while a base contains a OH- ion. From this definition, we can say that an acidic substance has a higher concentration of H+ ions. Now, I'll introduce here that pH is the value of the negative logarithm of the concentration of H+. In equation,
pH = -log[H+]
The term pOH is therefore also, pOH = -log[OH-]. Therefore, the relationship that connects the two negative logarithms is:
pH + pOH = 14
The pH scale starts from 1 being the most acidic to 14 being the most basic. The neutral pH is 7. Thus, for a pH of 7, the H+ and the OH- concentrations are equal. 
pH = 7 = -log[H+][H+] = 1×10⁻⁷ mol/L = [OH-]
Since the concentrations are equal, the ratio is equal to 1.

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fredd [130]

Explanation:

It is given that,

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Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

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F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

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Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

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6 0
2 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
Arlecino [84]

Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
2 years ago
Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as A to reach
aivan3 [116]
I believe this ratio is 4:1 due to the inverse square law
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2 years ago
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A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat
weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

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Therefore we can find the value of g at that location:

g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2

Now we can find the period of the second pendulum at the same location, which is given by

T=2\pi \sqrt{\frac{L}{g}}

where we have

L = 0.36 m (length of the  second pendulum)

g=17.5 m/s^2

Substituting,

T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s

#LearnwithBrainly

8 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

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the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

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Therefore, the internal energy of this gas is 433089.52 J.

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2 years ago
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