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Leno4ka [110]
3 years ago
7

How many liters of water vapor can be produced if 13.3 liters of methane gas (CH4) are combusted, if all measurements are taken

at 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:
Pie3 years ago
6 0
To get the number of liters of water vapor produced from the combustion of methane gas, we just need the stoichiometric ratio of water to methane which is 2:1. So the number of liters of water vapor from 13.3 liters of methane is 26.6 liters.
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Explanation:

It is given that two loads have 0.75 Ampere current each. And, they contain 2500 milli ampere per hour Ni-Cd battery.

As both the loads are connected in parallel. Hence, total current will be calculated as follows.

               I = I_{1} + I_{2}

                 = 0.75 A + 0.75 A

                 = 1.5 A

                 = 1.5 A \times \frac{1000 mA}{1 A}

                 = 1500 mA

Relation between time and capacity of battery is as follows.

             Capacity = Current × time (in hour)

therefore,        time = \frac{Capacity}{Current}

                                = \frac{2500 mA. h}{1500 A}

                                = 1.667 hr

Thus, we can conclude that the battery provide power to the load up to 1.667 hours.

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3 years ago
When an action force occurs, the reaction force is always Question 6 options: A.in the same direction as the action force. B.app
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D. 100%, extremely sure. Because Newton's 3rd law states every force has an equal and opposite reaction.
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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
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Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

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In order to prepare 10 mL, 5 μM from 25 μM solution,

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<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

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<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

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