Is rubber band stretching physical or chemical

Answers

Answer 1
Answer:

Answer:

\boxed {\boxed {\sf Physical \ change}}

Explanation:

A chemical change, as the name suggests, must result from a change in the chemical makeup and new substances are produced. A very common example is metal rusting.

A physical change can be observed without a change in chemical composition. No new substances are produces. Ripping paper and an ice cube melting are physical changes.

A rubber band stretching is a physical change. No new substances are made and the chemical makeup remains the same.

Answer 2
Answer:

Answer:

physical

Explanation:

this is a physical action to create a change. does not involve chemicals or elements


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The name of a water solution of hydrogen fluoride, HF(g), is (A) fluoric acid
(B) fluorous acid
(C) perfluoric acid
(D) hydrofluoric acid
(E) hypofluorous acid

Answers

the answer is D        asdasd

Most tornadoes carry wind speeds that range fromA. 1 to 105 km per hour.
B. 267 to 322 km per hour.
C. 105 to 177 km per hour.
D. 178 to 266 km per hour.

Answers

In the categorization of tornadoes, most tornadoes (87%) are no stronger than EF1 unless 'most' is considered 55% in which case the answer is EF0. 
EF0 - 65 to 85 mph {104.61 to 136.79 km/h}EF1 - 86 to 110 mph {138.40 km/h to 177 km/h}
Hence the answer would have to be C

Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(aq) v (b) io3-(aq) + fe2+(aq) equilibrium reaction arrow fe3+(aq) + i2(aq)

Answers

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Explanation:

We must look up the standard reduction potentials for the half-reactions.

                                                                    ℰ°    

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O     1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                       1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O             1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺                                      0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

                                                                                  ℰ°/V    

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻                  -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                                     1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻  + 6Cl⁻ + 14H⁺     0.00

(b) Fe²⁺/IO₃⁻

                                                                            ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻                                                -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O                      1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O     0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

Answer:

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Explanation:

We must look up the standard reduction potentials for the half-reactions.

                                                                   ℰ°    

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O     1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                       1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O             1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺                                      0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

                                                                                 ℰ°/V    

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻                  -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                                     1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻  + 6Cl⁻ + 14H⁺     0.00

(b) Fe²⁺/IO₃⁻

                                                                           ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻                                                -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O                      1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O     0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

Explanation:

its right trust

A chemist who is performing this reaction starts with 160.1 g of NH4NO3. The molar mass of NH4NO3 is 80.03 g/mol; the molar mass of water (H2O) is 18.01 g/mol. What mass, in grams, of H2O is produced?

Answers

the answer
the general equation of the reaction is 
NH4NO3 -------------- N2O   +  2H2O 

The molar mass of NH4NO3 is 80.03 
The mass of NH4NO3 is 160.1 g
 so we can deduce that  

n
NH4NO3 = number of mole of NH4NO3 = 160.1 g / 80.03  =2 moles
by using proportion property 

NH4NO3 -------------- N2O   +  2H2O 
1mole                       1mole     2 moles
nNH4NO3                                nH2O

the number of mole nH2O can be found by 

1/ nNH4NO3   =  2 /            nH2O 

           nH2O = 2 x  nNH4NO3 = 2 x 2 = 4 moles
to find the mass, we use 

mH2O = 4 moles x  18.01 g /mol = 36 g

Answer:

80.0

Explanation:

edge 2020

According to the quantum-mechanical model for the hydrogen atom, which electron transitions produces light with the longer wavelength: 2p→1s or 3p→1s?

Answers

Answer:2p→1s

Explanation:

Frequency is inversely proportional to wavelength. The higher the frequency, the shorter that wavelength. The smaller the frequency the longer the wavelength. The frequency of light associated with 2p→1s is less than the frequency of light associated with 3p→1s since the latter represents transition from a much higher energy level. It follows that the wavelength associated with the former will be longer since its frequency is lesser. This deduction follows from our argument about the inverse relationship between frequency and wavelength.

The reaction A→B has been experimentally determined to be second order. The initial rate is 0.0100M/s at an initial concentration of A of 0.300M. Determine the initial rate at [A]=0.900M.

Answers

Answer:

0.09 M/s is the initial rate when concentration of reactant A is 0.900 M.

Explanation:

A → B

1) Initial concentration of A = [A] =0.300 M

Rate constant of the reaction , k = ?

Rate equation for the second order kinetic is given as:

R=k[A]^2

0.0100M/s=k[0.300 M]^2

k=(0.0100M/s)/([0.300 M]^2)=0.1111 M^(-1) s^(-1)

2) Rate of the reaction when initial concentration of A was 0.900M be R'.

[A] = 0.900M

R'=k[A]^2

R'=0.1111 M^(-1) s^(-1)* [0.900 M]^2

R'=0.09 M/s