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STALIN [3.7K]
3 years ago
12

A sample of gas occupies 7.80 liters at 425°C? What will be the volume of the gas at 35°C if the pressure does not change?

Chemistry
1 answer:
balandron [24]3 years ago
4 0
From ideal gas equation PV = nRT, V/T = nR/P ==> V/T = constant. Therefore V1/T1 = V2/T2 ==> 7.8/698 = V2/308. V = 3.44L {TEMPERATURE IN KELVIN = 273 + 425 AND 35 = 698 AND 308}
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During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

                                    = 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0
3 years ago
Hellllpppppppp helppppppppp
mixas84 [53]

Answer :

Charles's Law : It is defined as the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

V\propto T

                                Boiling water bath        Cool bath 1       Cool bath 2

Temperature (⁰C)                  99                              17                       2

Temperature (K)(T)    273+99=372             273+17=290      273+2=275

Volume of water                  0.0                             27.0                34.0

in cool flask (mL)

Volume of water=              135.8                           135.8               135.8

Air in flask (mL)

Volume of air                    135.8                           108.8               101.8

in cool flask (V)

\frac{V}{T}                                \frac{135.8}{372}=0.365             \frac{108.8}{290}=0.375         \frac{101.8}{275}=0.370

The graph volume versus temperature for a gas is shown below.

3 0
3 years ago
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Answer:

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NARA [144]

Answer:

This question is incomplete

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