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lora16 [44]
3 years ago
7

An air/gasoline vapor mix in an automobile cylinder has an initial temperature of 180 ∘C and a volume of 13 cm3 . If the mixture

is heated to 587 ∘C with the pressure and amount held constant, what will be the final volume of the gas in cubic centimeters? Express your answer in cubic centimeters to three significant figures.
Chemistry
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

The final volume will be 24.7 cm³

Explanation:

<u>Step 1:</u> Data given:

Initial temperature = 180 °C

initial volume = 13 cm³ = 13 mL

The mixture is heated to a fina,l temperature of 587 °C

Pressure and amount = constant

<u>Step 2: </u>Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

V2 = (V1*T2)/T1

V2 = (13 mL *860 Kelvin) /453 Kelvin

V2 = 24.68 mL = 24.7 cm³

The final volume will be 24.7 cm³

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4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration
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Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100

<u><em>percentage composition of potassium iodide= 0.017%</em></u>

Finally, the percent composition of potassium iodide in this sample is 0.017%.

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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
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