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Contact [7]
3 years ago
13

Joshua hooks a rubber band between his thumb and forefinger. He moves his fingers apart, stretching the rubber band. With a quic

k, sudden motion, he bends his thumb forward so that the rubber band slips off. What happens next?
A. The chemical potential energy stored in Joshua's body changes to elastic potential energy in the rubber band.
B. The elastic potential energy stored in the rubber band changes to heat energy.
C. The chemical potential energy stored in Joshua's body changes to kinetic energy.
D. The elastic potential energy stored in the stretched rubber band changes to kinetic energy.
Chemistry
2 answers:
lana [24]3 years ago
8 0

Answer: D. The elastic potential energy stored in the stretched rubber band changes to kinetic energy.

Explanation: According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

Elastic potential energy is the energy stored as a result of applying a force to deform an elastic object. The energy is stored until the force is removed. When the force is removed,  the object gets back to its original shape.

Kinetic energy is the energy possessed by a body by virtue of its motion.

Gnoma [55]3 years ago
5 0
<span>The elastic potential energy stored in the stretched rubber band changes to kinetic energy.</span>
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Cu2 -&gt; Cu2+ is it getting oxidized?
KIM [24]

Answer:

Explanation:

Cu²⁺ + 2e⁻  →  Cu   ( copper gets reduced )

Cu  → Cu²⁺ + 2e⁻ (  copper gets oxidized )

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

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7 0
3 years ago
Discribe how to prepare 50.00mL of a 3.00M hydrochloric acid solution using 12.0 M HCl.
Leokris [45]

Answer:

The answer to your question is given below.

Explanation:

To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:

Molarity of stock solution (M1) = 12M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 3M

Volume of diluted solution (V2) = 50mL

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12 x V1 = 3 x 50

Divide both side by 12

V1 = (3 x 50)/12

V1 = 12.5mL

The volume of the stock solution needed is 12.5mL

Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.

6 0
3 years ago
A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

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          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

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