Answer:
The answer to your question is V2 = 1.82 l
Explanation:
Data
Volume 1 = 77 l
Pressure 1 = 18 mmHg
Volume 2 = ?
Pressure 2 = 760 mmHg
Process
Use Boyle's law to solve this problem
P1V1 = P2V2
-Solve for V2
V2 = P1V1/P2
-Substitution
V2 = (18 x 77) / 760
-Simplification
V2 = 1386 / 760
-Result
V2 = 1.82 l
12 times breathe give 240 ml of pure 
. Each breathe gives 20 ml of .
Let us consider, volume of air per breathe= x ml.
Pure from inhaled air= 
ml and Pure from exhaled air= 
ml.
Pure from inhaled and exhaled air= 20 ml
So, + = 20
Therefore, x = 55.5 ml
So, volume of air per breath= 55.5 ml.
N = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa, R = 8.314 L.kPa/K.mol
PV = nRT
V = nRT / P substituting.
V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (<span>101.325 kPa)
That is the answer, but it is not among the options you provided. Check your options properly.</span>
Line 1: straight horizontal line
Line 2: straight line at a slope
Line 3: exponential growth curve
Line 4: the topmost curve (the one that initially increases but then starts levels out)
Answer:
carbon dioxide
Explanation:
Carbon burns in oxygen to form carbon dioxide. Since hydrocarbon fuels only contain two elements, we always obtain the same two products when they burn. In the equation below methane (CH 4) is being burned. The oxygen will combine with the carbon and the hydrogen in the methane molecule to produce carbon dioxide (CO 2) and water (H 2O).
Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2. ... When the air or oxygen supply is restricted, incomplete combustion to carbon monoxide, CO, occurs. 2C(s) + O2(g) → 2CO(g) This reaction is important. When one mole of carbon is exposed to some energy in the presence of one mole of oxygen gas, one mole of carbon dioxide gas is produced. This reaction is a combustion reaction.
