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nikitadnepr [17]
3 years ago
14

please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi

ll there be in a room after 1 hour of time has passed?​
Chemistry
1 answer:
Nookie1986 [14]3 years ago
5 0

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>
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Pls help ty so much!
Nitella [24]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:mass = 11.42 \:\: grams

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: Q = ms\Delta T

  • \textsf{Q = heat evolved/absorbed = 501 J}

  • \textsf{m = mass in gram = ?}

  • \textsf{s = specific heat = 0.45}

  • \textsf{ΔT = change in temp = 120 - 22.5 =97.5°C}

\large\textsf{Find m : }

\qquad \tt \rightarrow \: 501 = m \sdot(0.45) \sdot(97.5)

\qquad \tt \rightarrow \:  501 = m \sdot43.875

\qquad \tt \rightarrow \: m =  \dfrac{501}{43.875}

\qquad \tt \rightarrow \: m  \approx11.42 \:  \: g

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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