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Ainat [17]
3 years ago
15

A mixture of propane and butane is fed into a furnace where it is mixed with air. the furnace exhaust leaves the furnace at 305°

c, 763.0 mmhg and contains only n2, o2, co2, and h2o. the partial pressure of o2 in the exhaust is 100.7 mmhg and the partial pressure of co2 in the exhaust is 33.57 mmhg.
Chemistry
1 answer:
IgorLugansk [536]3 years ago
3 0
In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.

   O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg)  = 0.13

   CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044

Answers: O2 = 0.13
               CO2 = 0.044
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what is the density of mercury if 205 mL has a mass of 2790 g; round your answer to the nearest 0.1 g/mL
kherson [118]
D = m / V

D = 2790 g / 205 mL

D = 13.60 g/mL
4 0
3 years ago
Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
3 years ago
If 63.8 grams of aluminum metal (Al) react with 72.3 grams of sulfur (S) in a synthesis reaction, how many grams of the excess r
Jobisdone [24]

Answer:

23.2 g of Al will be left over when the reaction is complete

Explanation:

2Al  +  3S  → Al₂S₃

1 mol of Al = 26.98 g

1 mol of S = 32.06 g

Mole = Mass / Molar mass

63.8 g/ 26.98 g/m = 2.36 mole of Al

72.3 g / 32.06 g/m = 2.25 mole of S

2 mole of Aluminun react with 3 mole of sulfur

2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S

As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.

3 mole of S react with 2 mole of Al

2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole

I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.

2.36 mole of Al - 1.50 mole of Al = 0.86 mole

This is the quantity of Al without reaction.

Molar mass . mole = Mass →  26.98 g/m . 0.86 m = 23.2 g

7 0
3 years ago
A runner runs 4339 ft in 7.45 minutes. What is the runner's average speed in miles per hou 6.62 mi/hr O 0.110 mi/hr 6.618 mi/h e
slavikrds [6]

<u>Answer:</u> The average speed of the runner is 6.618 miles/hr

<u>Explanation:</u>

Average speed is defined as the ratio of total distance traveled to the total time taken.

To calculate the average speed of the runner, we use the equation:

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

We are given:

Distance traveled = 4339 ft

Time taken = 7.45 mins

Putting values in above equation, we get:

\text{Average speed of runner}=\frac{4330ft}{7.45min}=582.42ft/min

To convert the speed into miles per hour, we use the conversion factors:

1 mile = 5280 ft

1 hr = 60 mins

Converting the speed into miles per hour, we get:

\Rightarrow \frac{528.42ft}{min}\times (\frac{1miles}{5280ft})\times (\frac{60min}{1hr})\\\\\Rightarrow 6.618mil/hr

Hence, the average speed of the runner is 6.618 miles/hr

6 0
3 years ago
At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i
olganol [36]

Answer:

K_c=0.0867

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.760}{1.50\ L}

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.130}{1.50\ L}

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}  

K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}

K_c=0.0867

5 0
3 years ago
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