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zheka24 [161]
3 years ago
9

The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?

Chemistry
2 answers:
nordsb [41]3 years ago
7 0
Mass of the gas m = 1.66 
The calculated temperature T = 273 + 20 = 293
 We have to calculate molar mass to determine the gas
 Molar Mass = mRT / PV
 M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
 M = 4043.76 / 101.3 = 39.92 g/mol
 So this gas has to be Argon Ar based on the molar mass.

PtichkaEL [24]3 years ago
6 0
Answer is: gas is argon (Ar).
Missing part of question find on internet:
part a: p(gas) = 1 atm, T(gas) = 20°C = 293,15 K.
part b: V(gas) = 1 L.
Ideal gas law: p·V = n·R·T or p·V = m/M·R·T.
R(universal gas constant) = 0,08206 L·atm<span>/mol·K.
</span>M = m·R·T÷p·V
M = 1,66 g·0,08206 L·atm<span>/mol·K</span>·293,15K÷(1 atm·1 L)
M = 39,9 g/mol (argon).

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Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

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Hence:

In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

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b)

here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

\dfrac{R_2}{R_1}= e^{0.00357}

\dfrac{R_2}{R_1}= 1.0035

where ;

R_2 = \dfrac{1}7{}

R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

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