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Dmitry_Shevchenko [17]
3 years ago
9

A mixture of propane and butane is fed into a furnace where it is mixed with air. The furnace exhaust leaves the furnace at 337�

C, 786.0 mmHg and contains only N2, O2, CO2, and H2O. The partial pressure of O2 in the exhaust is 10.38 mmHg and the partial pressure of CO2 in the exhaust is 88.03 mmHg.
1.What is the mole fraction of propane in the fuel stream?
2.What is the mole fraction of water in the exhaust stream?
3.What is the dew point temperature of the exhaust gas?

Chemistry
2 answers:
vagabundo [1.1K]3 years ago
7 0

Answer:

1. 3.29mol

2. 0.125molH2o/mol

3. 52.5'C

Explanation: The step by step explanation are attached to the answer.

babymother [125]3 years ago
3 0

Answer:

Mole fraction of propane = 0.74 mol C₃H₈/mol

Mole fraction of water = 0.29 mol H₂0 /mol

Dew point temperature = 52.5°C

Explanation:

See the attached file for the calculation.

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Gravity pulls gas and dust together. A protostar forms as mass increases.
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What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?
galina1969 [7]

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for  0.00135 moles

= 2 ×  0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

or

Volume =  0.0303 Liters

8 0
3 years ago
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A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of
Aleksandr-060686 [28]

<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0
3 years ago
At a certain temperature and pressure, one liter of CO2 gas weighs 1.95 g.
AysviL [449]

Answer:

1.332 g.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):

<em>∴ (n) of CO₂ = (n) of C₂H₆</em>

<em></em>

∵ n = mass/molar mass

<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>

mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.

mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.

<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>

<em></em>

7 0
3 years ago
Hey can someone pls help me answer these questions? It should be easy maybe. It’s just science. The question just won’t to know
miskamm [114]
Hehehehwgwgw. Be the hardest thing ever for a long day and I have a windows
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