Answer: Physical change : tearing of paper, fixing of wtaer
Chemical change: rusting of iron , electrolysis of water, Rancidification
Explanation:
Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.
Example: tearing of paper, fixing of wtaer
Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.
Example: rusting of iron , electrolysis of water, Rancidification
The reactions are in order which includes combustion reaction, Hydration reaction, oxidation reaction, and displacement reaction.
a) A combustion reaction is a chemical reaction between a fuel and an oxidant where heat is released. The combustion reaction example is given below. It is a balanced chemical reaction.
2C₃H₆(g) + 9O₂(g) --------> 6CO₂(g) + 6H₂O(g)
b. A hydration reaction is a chemical reaction in which a molecule of water is added to another molecule. Here Aluminum oxide is added to water to form aluminum hydroxide.
4Al₂O3(s) + 6H₂O(l)------> 2Al(OH)3(s)
c. When a metal reacts with oxygen, the metal forms an oxide. Oxide is a compound of metal and oxygen. Here lithium metal reacts with oxygen to form lithium oxide.
2Li(s) + O₂(g)-----> Li₂O(s)
d. A displacement reaction is one in which a more reactive element displaces a less reactive element from a compound. Here Zinc is more reactive than silver, so silver was displaced to form Zinc Nitrate.
Zn(s) + 2AgNO₃(aq) -----> 2Ag(s) + Zn(NO₃)₂(aq)
To know more about reactions, click below:
brainly.com/question/11231920
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Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
