Mostly because the lake smell really bad and the waters really dirty
Le us discuss how we should approach this problem. We know that there are two states. The first state is when P₁ = 1 bar and V₁ = 1.59 L. The second state is when P₂ = 80 bar. We are asked to find V₂. At constant temperature, we use Boyle's Law.
P₁V₁ = P₂V₂
V₂ = P₁V₁/P₂ = (1 bar)(1.59 L)/(80 bar)
<em>V₂ = 0.02 L</em>
Answer:
The devise used to convert thermal energy to mechanical energy are called heat enignes
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94.0 L if the reaction takes place under STP.
<h3>Explanation</h3>
The molar mass of glucose C₆H₁₂O₆ is
12.01 × 6 + 1.008 × 12 + 16.00 × 16.00 = 180.16 g / mol.
126 grams of glucose will contain 126 / 180.16 = 0.69939 mol of C₆H₁₂O₆. (To avoid rounding errors, keep a couple more digits than necessary.)
6 moles of CO₂ will be produced when 1 mole of C₆H₁₂O₆ is consumed. 0.69939 moles of C₆H₁₂O₆ will give rise to 4.196 mol of CO₂.
Assuming that the reaction takes place under STP, where T = 0 °C = 273 K and P = 1 atm. Each mole of any ideal gas will occupy a volume of 22.4 liters. The 4.196 moles of CO₂ will occupy 4.196 × 22.4 = 94.0 L. (The least significant number given is 126 g, the mass of glucose. This number has three significant figures. Thus, round the result to three significant figures.)
The volume of CO₂ can be found using the ideal gas law if the condition isn't STP. For example, T = 25 °C = 297 K and P = 1.00 × 10⁵ will lead to a different volume. By the ideal gas law,
V = (n · R · T) / (P)
where
- V is the volume of the gas,
- n is the number of moles of gas particles,
- R is the ideal gas constant<em>,</em>
- P is the pressure on the gas,
- T is the absolute temperature of the gas (in degrees Kelvin.)
R = 8.314 × 10³ L · Pa / (K · mol)
Taking T = 297 K and P = 1.00 × 10⁵ Pa,
V = (4.196 × 8.314 × 10³ × 297 ) / (1.00 × 10⁵ ) = 104 L.
