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Andrews [41]
3 years ago
13

Which scenarios are examples of physical changes? Check all that apply.

Chemistry
2 answers:
lawyer [7]3 years ago
8 0

Explanation:

Physical change is a change that does not affect the chemical composition of a substance but it changes the physical appearance like size or shape of the substance.

For example, crushing a can, melting of ice, breaking a glass etc represents the physical change.

Scenarios which represent examples of physical changes are as follows.

  • Silicon is cut into tiny pieces for computer chips.
  • Lead is melted to form pellets.
  • Liquid water freezes to form ice.


Alona [7]3 years ago
3 0

Answer:

Lead is melted to form pellets

Explanation:

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The equation to be used are:

PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles

The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.

PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa

n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
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7 0
3 years ago
Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces
inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt  

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

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By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>    

aX → bY (1)

rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}

<em>where, a and b are the coefficients of de reactant X and product Y, respectively.        </em>

(a) Based on the definition above, we can express the rate of reaction (2) as follows:      

3O₂(g) → 2O₃(g) (2)    

rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t} (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:  

rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}

+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}          

\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

           

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6 0
4 years ago
The octet rule is the basis for the stability of the atom. According to this rule, the outermost shell of the atom must contain
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Answer:

the answer would be 0.666 mL

Explanation:

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