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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Answer:
A table was attached to the question
Explanation:
The step by step calculation is as shown
The balanced chemical reactions are:
Further Explanation:
The following reactions will undergo double displacement where the metal cations in each compound are exchanged and form new products.
For reaction 1, the compounds involved are nitrates and chlorides. To determine the states of the products, the solubility rules for nitrates and chlorides must be followed:
- All nitrates are generally soluble.
- Chlorides are generally soluble except AgCl, PbCl2, and Hg2Cl2.
Therefore, the products will have the following characteristics:
- silver chloride (AgCl) is insoluble
- sodium nitrate (NaNO3) is soluble
For reaction 2, the compounds involved are phosphates and chlorides. The solubility rules for phosphates and chlorides are as follows:
- Phosphates are generally insoluble except for Group 1 phosphates.
- Chlorides are generally soluble except for AgCl, PbCl2, and Hg2Cl2.
Hence, the products of the second reaction will have the following characteristics:
- potassium chloride (KCl) is soluble
- magnesium phosphate is insoluble
Insoluble substances are denoted by the symbols (s) in a chemical equation. The soluble substances are denoted as <em>(aq).</em>
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Keywords: solubility rules, precipitation reaction