The combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how much oxygen was used up in grams
1 answer:
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Density is defined as the ratio of mass to the volume.
Density = (1)
Mass of water = 10 grams
Mass of acetone = 10 grams
Density of water = 1
Density of acetone = 0.7857
Put the value of density of water and its mass in equation (1)
1 =
Volume of water = 10
Put the value of density of acetone and its mass in equation (1)
0.7857 =
Volume of acetone = 12.72
Thus, volume of acetone is more than volume of water because the density of acetone is lower.
Answer:
Explanation:
The gas ideal law is
PV= nRT (equation 1)
Where:
P = pressure
R = gas constant
T = temperature
n= moles of substance
V = volume
Working with equation 1 we can get
The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.
or
(equation 2)
The cylindrical container has a constant pressure p
The volume is the volume of a cylinder this is
Where:
r = radius
h = height
(pi) = number pi (3.1415)
This cylinder has a radius, r and height, h so the volume is
Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:
Replacing these values in the equation 2 we get:
(equation 2)
Answer:
2.42L
Explanation:
Given parameters:
V₁ = 1.8L
T₁ = 293K
P₁ = 101.3kPa
P₂ = 67.6kPa
T₂ = 263K
Unknown:
V₂ = ?
Solution:
To solve this problem, we are going to use the combined gas law to find the final volume of the gas. The combined gas law expression combines the equation of Boyle's law, Charles's law and Avogadro's law;
All the units are in the appropriate form. We just substitute and solve for the unknown;
101.3 x 1.8 / 293 = 67.6 x V₂ / 263
V₂ = 2.42L