The number of moles of C present in C₅H₁₂ that contains 22.5 g of H is 9.375 moles
<h3>How to determine the mass of C₅H₁₂ that contains 22.5 g of H</h3>
1 mole of C₅H₁₂ = (12×5) + (1×12) = 72 g
Mass of H in 1 mole of C₅H₁₂ = 12 × 1 = 12 g
Thus,
12 g of H is present in 72 g of C₅H₁₂
Therefore,
22.5 g of H will be present in = (22.5 × 72) / 12 = 135 g of C₅H₁₂
<h3>How to determine the mole of C present in 135 g of C₅H₁₂</h3>
72 g of C₅H₁₂ contains 5 moles of C
Therefore,
135 g of C₅H₁₂ will contain = (135 × 5) / 72 = 9.375 moles of C
Thus, 9.375 moles of C is present in C₅H₁₂ that contains 22.5 g of H
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When making a solution by diluting a stock solution. Suppose when a solid is stirred into a liquid and dissolves,
a homogenous mixture is formed. When the mixture does not dissolve, the mixture
is heterogeneous.
Density is an intensive property of a substance. An intensive property means that it is a physical property that does not depend on the size or the amount of material in the matter. For example, if a diamond is cut in half, the density does NOT change, therefore it is an intensive property.
Answer:
Mass = 2.355 g
Explanation:
Given data:
Mass of K₂O needed = ?
Mass of KNO₃ produced = 5.00 g
Solution:
Chemical equation:
K₂O + Ca(NO₃)₂ → CaO + 2KNO₃
Number of moles of KNO₃:
Number of moles = mass/molar mass
Number of moles = 5.00 g/ 101.1 g/mol
Number of moles = 0.05 mol
now we will compare the moles of KNO₃ and K₂O.
KNO₃ : K₂O
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of potassium oxide needed in gram:
Mass = number of moles × molar mass
Mass = 0.025 mol × 94.2 g/mol
Mass = 2.355 g
No, the two isotopes of lithium-6 and lithium-7 are not equally common.
The more plentiful isotope would be lithium-7.
This can be easily demonstrated by assuming that both isotopes were equally common. If that were the case, the average atomic mass would be (6 + 7)/2 = 6.5 amu. Now compare that value if they were both equal to the actual value found in nature. The value found in nature is 6.941 amu which is heavier than the 6.5 amu that would happen if they were equally common. Since the natural value is heavier, that means that there has to be more of the heavier isotope than there is of the lighter one. Therefore lithium-7 is more common than lithium-6.