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anygoal [31]
3 years ago
7

Jack is playing with a Newton's cradle. As he lifts one ball to position A and drops it, it impacts the other balls at position

B and becomes stationary while the ball on the opposite side leaps into the air swinging to position C. What is happening to cause this?
Chemistry
1 answer:
svetlana [45]3 years ago
8 0

Answer:

Newton's Cradle is a neat way to demonstrate the principle of the CONSERVATION OF MOMENTUM.

What happens here is when the ball on one end of the cradle is swung and it hits the other balls that are motionless, or stationary, the momentum of the swinging ball is transferred to the next ball upon impact.

Momentum is not lost in this action, what happens when it hits the next ball, the momentum is transferred to the next one, and then the next, and the the next, till it reaches the last ball on the other end. Since nothing is next to the last ball, it pushes the ball upwards, which will swing down and repeat the process going the other way.

This also demonstrates the CONSERVATION OF ENERGY. As you will see, the energy continues to move through the other balls, passing it from one ball to the other, which keeps this constantly moving.

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In your own words, explain how multiple approaches to a scientific investigation can be used to obtain the same result. Provide
aleksandr82 [10.1K]

Answer:B

Explanation:

5 0
3 years ago
What is the volume of 0.4 mole of gas at stp? 30 points!
iragen [17]

Answer:

89.6 liters

Explanation:

A STP (standard temperature and pressure) ONE mole of any idea gas will occupies 22.4 liters. So,...

4 moles x 22.4 L/mol = 89.6 liters

5 0
2 years ago
Read 2 more answers
Write both answers to at least two decimal places. Calculate the pH of a 0.160 M solution of KOH.Part 2 (1 point) Calculate the
Evgen [1.6K]

To calculate the pH of a solution, we first need to find the concentration of hydronium ions in the solution. Since KOH is a strong base, it dissociates completely in water to produce hydroxide ions (OH-) and potassium ions (K+).

The concentration of hydronium ions in a solution of KOH can be calculated using the concentration of hydroxide ions and the equilibrium constant for water, which is equal to 1.00 x 10^-14 at 25 degrees Celsius.

The concentration of hydroxide ions in a 0.160 M solution of KOH is equal to the concentration of KOH, which is 0.160 M. The concentration of hydronium ions in the solution can be calculated using the equation below:

[H3O+] = (1.00 x 10^-14) / [OH-]

Substituting the concentration of hydroxide ions into the equation above, we get:

[H3O+] = (1.00 x 10^-14) / (0.160 M) = 6.25 x 10^-13 M

To calculate the pH of the solution, we need to take the negative logarithm of the concentration of hydronium ions. This can be done using the equation below:

pH = -log([H3O+])

Substituting the concentration of hydronium ions into the equation above, we get:

pH = -log(6.25 x 10^-13) = 12.20

The pH of a 0.160 M solution of KOH is 12.20.

To calculate the pOH of a solution, we first need to find the concentration of hydroxide ions in the solution. Since we already calculated this value above, we can simply use the concentration of hydroxide ions we found earlier: 0.160 M.

To calculate the pOH of the solution, we need to take the negative logarithm of the concentration of hydroxide ions. This can be done using the equation below:

pOH = -log([OH-])

Substituting the concentration of hydroxide ions into the equation above, we get:

pOH = -log(0.160 M) = 1.80

The pOH of a 0.160 M solution of KOH is 1.80.

Learn more about pH:
brainly.com/question/28864035

#SPJ4

7 0
1 year ago
What is the molarity of a solution prepared by diluting 43.72 ml of 1.005 M aqueous K2CR2O7 to 500 ml
UNO [17]

Answer:

43.72

Explanation:

that is the answer hope u liked it and I did this already along time ago

3 0
3 years ago
PH CHEM, PLEASE HELP QUICK! NO LINKS/VIRUSES PLEASE!
Contact [7]

Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH  +  HCl  →  H₂O  +  NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles

Now, we need to find the concentration of the OH⁻ ions.

[OH^{-}] = \frac{n_{NaOH}}{V}

Where V is the volume of the solution = 1.00 L                

[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L

Finally, we can calculate the pH of the solution as follows:

pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52

pH + pOH = 14

pH = 14 - pOH = 14 - 2.52 = 11.48

Therefore, the pH of the solution is 11.48.

I hope it helps you!

3 0
3 years ago
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