Physical change because it's only changing the shape of the gold.
Answer: 41.46 L
Explanation:
La ecuación que describe relación entre presión, volumen, temperatura y la cantidad (en moles)
de un gas ideal es:
PV = nRT
Donde: P = Presión absoluta
, V= Volumen , n = Moles de gas
, R = Constante universal de los gases ideales, T = Temperatura absoluta,
R = 0.082 L. atm/mol. °K
V = nRT/P
Calculanting n
n = mass/ molecular mass
<h3>n = 4 g / 2g. mol⁻¹</h3><h3>n = 2 mol</h3><h3>T =25⁰ + 273 ⁰K = 298 ⁰K</h3><h3>V = (2 mol ₓ0.082 L. atm / mol.°K x 298 ⁰K) / 1.18 atm = 41.46 L</h3>
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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The correct answer would b C
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg
