Acacia ant and acacias animals are an example of coevolution
Coevolution is the reciprocal evolutionary change in a set of interacting population over time resulting from the interaction between those population and an example of coevolution that is not characteristics of an arm race but one which provides a mutual benefit to both a plant species and insect is that of the acacia ant and acacia plant and many cases of coevolution can be found between plants and insects
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Magnetism: separates magnetic materials from non-magnetic. 3rd diagram
evaporation: separates a soluble solid by boiling off. 4th diagram
filtration: separates insolube solid from a liquid. 1st diagram
distillation: separates liquids with different boiling points. 5th diagram
chromatography: separates liquids of different colours. 2nd diagram
1. two or mor, chemically bonded
2. purify
3. magnetism, evaporation, filtration, distillation, chromatography
The answer is Photosphere Apex.
Answer:
Answers with detail are given below
Explanation:
1) Given data:
Mass of Rb₃Rn = 76.19 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 478.43 g/mol
Number of moles = 76.19 g/ 478.43 g/mol
Number of moles = 0.16 mol
2) Given data:
Mass of FrBi₂ = 120.02 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 640.96 g/mol
Number of moles = 120.02 g/640.96 g/mol
Number of moles = 0.19 mol
3) Given data:
Mass of Zn₂F₃ = 88.24 g
Number of moles = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 187.73 g/mol
Number of moles = 88.24 g/ 187.73 g/mol
Number of moles = 0.47 mol
4) Given data:
Number of moles of Sb₄Cl = 1.20 mol
Mass of Sb₄Cl = ?
Solution:
Number of moles = mass/molar mass
Molar mass = 522.49 g/mol
Mass = Number of moles × molar mass
Mass = 1.20 mol × 522.49 g/mol
Mass = 626.99 g
The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
