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ser-zykov [4K]
2 years ago
14

Convert 338 L at 63.0 atm to its new volume at standard pressure.

Chemistry
1 answer:
taurus [48]2 years ago
7 0

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = \frac{P1V1}{P2}

Putting the values in the equation:

V2 = \frac{338X63}{1}

     = 21294 L

as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

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A sample of fluorine gas is confined in a 5.0-L container at 0.432 atm and 37 °C. How many moles of gas are in this sample?
Ahat [919]

Answer:

About 0.08486 moles

Explanation:

PV=nRT, when P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

0.432 \cdot 5= n \cdot 0.0821 \cdot 310

n\approx 0.08486 moles

Hope this helps!

7 0
2 years ago
Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the wa
Kryger [21]

Explanation:

As the given reaction is as follows.

     H_{2}S(aq) + Cl_{2}(aq) \rightarrow S(s) + 2H^{+}(aq) + 2Cl^{-}(aq)

So, according to the balanced equation, it can be seen that rate of formation of Cl^{-} will be twice the rate of disappearance of H_{2}S .

And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.

This means that,

Rate of the reaction = -Rate of disappearance of H_{2}S

                 = k[H_{2}S][Cl_{2}]

                 = (3.5 \times 10^{-2}) \times (2 \times 10^{-4}) \times (2.8 x 10^{-2})

                 = 1.96 \times 10^{-7} M/s

Therefore, calculate the rate of formation of Cl^{-} as follows.

Rate of formation of Cl^{-} = 2 \times 1.96 \times 10^{-7}

                                        = 3.92 \times 10^{-7} M/s

Thus, we can conclude that the rate of formation of Cl^{-} is 3.92 \times 10^{-7} M/s.

5 0
2 years ago
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bee
lora16 [44]

Answer:

The answer to your question is:

a) 4.64 x 10 ¹⁵ molecules

b) 9.28 x 10 ¹⁵ atoms of O2

Explanation:

MW C7H14O2 = 84 + 14 + 32 = 130 g

a)        130 g of C7H14O2 ---------------- 1 mol of C7H14O2

           1 x 10 ⁻⁶ g              ---------------      x

           x = 7.7 x 10 ⁻⁹ mol

          1 mol of C7H14O2   --------------   6 .023 x 10 ²³ molecules

          7.7 x 10⁻⁹ mol          --------------    x

          x = 4.64 x 10¹⁵ molecules

b)      130 g of C7H14O2   ----------------   1 mol of C7H14O2

         1 x 10⁻⁶  C7H14O2   -----------------     x

         x = 7.7 x 10 ⁻⁹ mol of C7H14O2

        1 mol of C7H14O2    ---------------   2 mol of O2

        7.7 x 10 ⁻⁹                 ----------------   x

         x = 1.54 x 10⁻⁸ mol of O2

       1 mol of O2 -----------------  6.023 x 10 ²³ atoms

       1.54 x 10 ⁻⁸  ----------------   x

        x = 9.28 x 10 ¹⁵ atoms of O2

8 0
2 years ago
Brian's aunt has cats. When Brian recently visited her, he started sneezing badly and believes that it was because
kifflom [539]

Answer:

By visiting other households with cats.

Explanation:

This will give Brian a variety of other houses and determine if it is truly cats or just alleries from other items. This is the most direct way to get Brian the answer he is looking for.

4 0
2 years ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
2 years ago
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