If your talking about physical property and chemical change then , a physical property is something that can be observed without changing and a chemical change , changes into a new substance.
Given ,
Mass of sample of cobalt = 27 g
density of sample of cobalt = 9g/cm^3
We know that ,
Density = mass of sample/volume of sample
From that relation ,
We can deduce the following as
Volume = mass of sample/density of sample
Hence , required volume of sample of cobalt = 27 g /9 g/cm^3 = 3 cm^3
The volume is
2.1653 g
Explanation:
The molar mass of Rubidium is;
85.468 g/mol
Therefore the moles of Rubidium that reacted with oxygen is;
1.98 / 85.468
= 0.0232 moles
If every two moles of Rubidium reacts with one mole of oxygen then the amount of oxygen consumed in the chemical reaction is;
0.5 * 0.0232
= 0.0116 moles
The molar mass of an oxygen atom is 16 g/mole. Then the amount of O in grams consumed is;
0.0116 * 16
=0.1853 g
The final weight of the Rubidium II Oxide is;
1.98 + 0.1853
= 2.1653 g
Answer:
20.95 g of caffeine, C₈H₁₀N₄O₂
Explanation:
From the question given above, the following data were obtained:
Number of molecules of C₈H₁₀N₄O₂ = 6.5×10²² molecules
Mass of C₈H₁₀N₄O₂ =?
From Avogadro's hypothesis,
1 mole of C₈H₁₀N₄O₂ = 6.02×10²³ molecules
Next, we shall determine the mass of 1 mole of C₈H₁₀N₄O₂. This can be obtained as follow:
1 mole of C₈H₁₀N₄O₂ = (8×12) + (10×1) + (4×14) + (2×16)
= 96 + 10 + 56 + 32
1 mole of C₈H₁₀N₄O₂ = 194 g
Thus,
194 g of C₈H₁₀N₄O₂ = 6.02×10²³ molecules
Finally, we shall determine the mass of caffeine, C₈H₁₀N₄O₂ that contains 6.5×10²² molecules. This can be obtained as follow:
6.02×10²³ molecules = 194 g of C₈H₁₀N₄O₂
Therefore,
6.5×10²² molecules = (6.5×10²² × 194) / 6.02×10²³
6.5×10²² molecules = 20.95 g of C₈H₁₀N₄O₂.
Therefore, 20.95 g of caffeine, C₈H₁₀N₄O₂ contains 6.5×10²² molecules
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