Which starting condition would produce the larger volume of carbon dioxide by combustion of ch4(g) with an excess of oxygen gas to produce carbon dioxide and water: (a) 2.00 l of ch4(g); (b) 2.00 g of ch4(g)? justify your answer. the system is maintained at a temperature of 75 c and 1.00 atm.

(b) 2.00 g of CH4(g)

Explanation:

(a) 2.00 L of

Volume = 2.00 L

Temperature = 75°C

Pressure = 1.00 atm

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (75 + 273.15) K = 348.15 K  

T = 348.15 K

Using ideal gas equation as:

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1.00 atm × 2.00 L = n × 0.0821 L.atm/K.mol ×  348.15 K

⇒n = 0.07 moles

(b) 2.00 g of

Mass of = 2.00 g

Molar mass of = 16.04 g/mol

The formula for the calculation of moles is shown below:

Thus,

Moles of = 0.125 moles

Thus, more moles, more volume of carbon dioxide will be produced when more moles of methane react.

So, Answer - (b) 2.00 g of CH4(g)

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

unbalanced force

explanation: