<h3>
Answer:</h3>
8CO₂
<h3>
Explanation:</h3>
We are given;
- Butane is a hydrocarbon in the homologous series known as alkane.
We are required to determine the other product produced in the combustion of butane apart from water.
- We know that the complete combustion of alkane yields carbon dioxide and water.
- Therefore, combustion of butane will yield carbon dioxide and water.
- The balanced equation for the complete combustion of butane will be;
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
Answer: 167 g
Explanation:
1) The depression of the freezing point of a solution is a colligative property ruled by this equation:
ΔTf = i × m × Kf
Where:
ΔTf is the decrease of the freezing point of the solvent due to the presence of the solute.
i is the Van't Hoof factor and is equal to the number of ions per each mole of solute. It is only valid for ionic compounds. Here the solute is not ionice, so you take i = 1
Kf is the molal freezing constant and is different for each solvent. For water it is 1.86 m/°C
2) Calculate the molality (m) of the solution
ΔTf = i × m × Kf ⇒ m = ΔTf / ( i × Kf) = 5.00°C / 1.86°C/m = 2.69 m
3) Calculate the number of moles from the molality definition
m = moles of solute / kg of solvent ⇒ moles of solute = m × kg of solvent
moles of solute = 2.69 m × 1.00 kg = 2.69 moles
4) Convert moles to grams using the molar mass
molar mass of C₂H₆O₂ = 62.07 g/mol
mass in grams = number of moles × molar mass = 2.69 moles × 62.07 g/mol = 166.97 g ≈ 167 g
