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Sidana [21]
3 years ago
12

How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken a

t the same temperature and pressure? Show all of the work used to solve this problem.
CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)
Chemistry
1 answer:
Delvig [45]3 years ago
8 0
We know, mole of gas in ideal conditions is 22.414 L
Here, CH4 volume = 8.9L

So, number of moles = 8.9 / 22.414 = 0.397 moles
Here, in balanced equation we have a ratio CH4 : H2O = 1:2.

So, mole so water = , 0.397*2= 0.794 moles
In liters it would be: 0.794 * 22.414 = 17.80 L

In short, Your Answer would be 17.80 Liters

Hope this helps!
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Herbie left his house at 3:30 p.m. and arrived at Kit’s house at 5:00 p.m. The two boys live 7 miles apart. What was Herbie’s av
Kisachek [45]

Answer:

Step-by-step explanation:

Alright, lets get started.

Suppose they take t minutes to meet each other.

Distance covered by first friend in t minutes, = 0.2 *t=0.2∗t

Distance covered by second friend in t minutes , =0.15 *t=0.15∗t

Total distance is given as 7, so

0.2 t + 0.15 t = 70.2t+0.15t=7

0.35 t = 70.35t=7

t = 20t=20

means after 20minutes they will meet.

SO. the average speed is 10m: Answer

5 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
What is a chocking smell​
Goshia [24]

Answer:

A horrible, nasty smell.

Explanation:

A choking smell is a nasty, horrible smell. Hope this helps!

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3 years ago
Which is true about covalent compounds?
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Covalent compounds have a low boiling point. All the other answers apply to ionic compounds.
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