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gayaneshka [121]
3 years ago
6

A baseball is traveling (+20 m/s) and is hit by a bat. It leaves the bat traveling (-30 m/s). What is the change in the velocity

?
A) 10 m/s

B) 20 m/s

C) 30 m/s

D) 50 m/s
Chemistry
2 answers:
zepelin [54]3 years ago
7 0

Answer: Option (D) is the correct answer.

Explanation:

Change in velocity is calculated when initial velocity is subtracted from final velocity.

Formula to calculate change in velocity is as follows.

            \Delta V = final velocity (V_{f}) - initial velocity (V_{i})

                                  = (-30 m/s) - (+20 m/s)

                                  = (-30 - 20) m/s

                                  = - 50 m/s

Therefore, change in velocity is 50 m/s.

Tpy6a [65]3 years ago
4 0
D.

A more accurate answer would be -50 m/s. 50 m/s is the change in speed.
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C) Has sub levels within the energy levels. Lol forgot to put C
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Consider the electrolysis of molten barium chloride, BaCl2. (a) Write the half-reactions. (b) How many grams of barium metal can
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Answer: a)  Cathode(-): Ba^+^2+2e^-\rightarrow Ba

Anode(+): 2Cl^-\rightarrow Cl_2+2e^-

b) 0.640 grams of Ba will be deposited.

Explanation:  a) The problem is based on Faraday law of electrolysis. Molten barium chloride has Ba^+^2 ion and Cl^- ion. Barium ion is reduced and chloride is oxidized. Reduction takes place at cathode and oxidation at anode. So, the half reactions will be:

Cathode(-): Ba^+^2+2e^-\rightarrow Ba

Anode(+): 2Cl^-\rightarrow Cl_2+2e^-

b) The question asks, how many grams of barium metal can be produced by supplying 0.50 ampere for 30 minutes.

From the Cathode half-reaction, 1 mole of Ba is deposited by 2 moles of electrons and we know that 1 mole of electron carries one Faraday that is 96485 Coulomb.

Coulombs for 2 moles of electrons will be = 2*96485 C = 192970 C

So, we can say that, 192970 C will deposit 1 mole of Ba metal.

Total available coulombs can be calculated using the formula:

q=i*t

where, q is electric charge in coulomb, i is current in ampere and t is time in seconds.

q=q=0.50A*30min(\frac{60sec}{1min})

q = 900 C         (note: 1 C = 1 A*sec)

Let's calculate how many moles of Ba will get deposited by 900 C.

900C(\frac{1molBa}{192970C})

= 0.00466 mole Ba

Convert the moles of Ba to grams and for this we multiply by molar mass of Ba which is 137.33 gram per mol.

0.00466molBa(\frac{137.33g}{1mol})

= 0.640 g Ba

So, 0.50 A for 30 minutes will deposit 0.640 grams of Ba metal.

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Explanation:

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