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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1.) Calculate the change in pH when 4.00 ml of 0.100 M HCl(aq) is added to 0.100 mL of a buffer solution that is 0.100 M in NH3 (aq) and 0.1000 M in NH4Cl(aq).
2.) Calculate the change in pH when 4.00 mL of 0.100 M in NaOH(aq) is added to the original buffer solution.
Answer is: 4) The same subscripts are on each side of the equation.
For example, balanced chemical reaction:
2Mg + O₂ → 2MgO.
1) The same number of atoms is on each side of the equation: two magnesium atoms and two oxgen atoms.
2) The formulas for all substances are correct: in magnesium oxide (MgO), magnesium has oxidation number +2 and oxygen -2, so formula is good, because compound must be neutral.
3) The same mass is represented on each side of the equation: because there is same number of atoms, the mass is the same.
4) The same subscripts are on each side of the equation: oxygen does not have same subscripts.
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