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Gala2k [10]
3 years ago
6

What is the mass, in pounds, of 389 mL of a gas that has a density of 1.29 g/L?

Chemistry
1 answer:
devlian [24]3 years ago
7 0

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of gas = ?

Density of gas = 1.29g/L

Volume of gas = 389 ml = 0.389 L     (1L=1000ml)

Putting in the values we get:

1.29g/L=\frac{mass}{0.389L}

mass=0.5grams  

mass=0.5\times 0.002lb=0.001lb       (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

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Explanation:

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1.36g H2 is allowed to react with 10.1g N2 producing 2.05g NH3 What is the theoretical yield in grams for this reaction under th
vladimir2022 [97]

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3 years ago
3.2 moles of H3PO4 to grams
Nimfa-mama [501]

Answer:

313, 6grams of H3PO4

Explanation:

We calculate the weight of 1 mol of H3PO4:

Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4  =1gx3+31g+16gx4

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1 mol-----98 grams H3PO4

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4 0
3 years ago
Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer. Trichlorofluoromethane (CCl₃
Daniel [21]

Answer:

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

Explanation:

Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

S=\frac{Q}{T(Kelvins)}

Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol

Temperature at which Trichlorofluoromethan boils , T= 296.95 K

The molar entropy of the evaporation of Trichlorofluoromethan :

=\frac{24.8 kJ/mol}{296.95 K}=0.083516 kJ/mol K = 83.516 J/molK

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

8 0
3 years ago
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
vazorg [7]

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

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

Hence:

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

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

16.1782126= Ea * 9.0292*10^{-5}

Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

Ea = 179.176 kJ/mol

Thus; the activation energy Ea = 179.176 kJ/mol

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.

4 0
3 years ago
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