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Shtirlitz [24]
3 years ago
10

Question 7 of 15

Chemistry
1 answer:
Crazy boy [7]3 years ago
3 0

Answer: 0.4 moles

Explanation:

Given that:

Volume of gas V = 11L

(since 1 liter = 1dm3

11L = 11dm3)

Temperature T = 25°C

Convert Celsius to Kelvin

(25°C + 273 = 298K)

Pressure P = 0.868 atm

Number of moles N = ?

Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)

9.548 atm dm3 = n x 24.47atm dm3mol-1

n = (9.548 atm dm3 / 24.47atm dm3 mol-1)

n = 0.4 moles

Thus, there are 0.4 moles of the gas.

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A researcher was attempting to quantify the amount of dichlorodiphenyltrichloroethane (DDT) in spinach with gas chromatography u
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Answer:

0.0136mg DDT / g spinach

Explanation:

Quantification in chromatography by internal standard has as formula:

RF = Aanalyte×Cstd / Astd×Canalyte <em>(1)</em>

<em>Where RF is response factor, A is area and C is concentration</em>

Replacing with first experiment values:

RF = 5019×3.20mg/L / 8179×6.37mg/L

RF = 0.308

In the next experiment, final concentration of chloroform was:

11.45mg/L × (1.25mL / 25.00mL) = <em>0.5725mg/L</em>

From (1), it is possible to write:

Aanalyte×Cstd / Astd×RF = Canalyte

Replacing:

6821×0.5725mg/L / 14061×0.308 = Canalyte

Canalyte = <em>0.9017mg/L</em>

as the sample was made from 0.750mL of extract. Concentration of extract is:

0.9017mg/L × (25.00mL / 0.750mL) = 30.06mg/L. As the extract has a volume of 2.40mL:

30.06mg/L × 2.40x10⁻³L = <em>0.07213mg of DDT in the extract</em>

As the extract was made from 5.29g of spinach:

0.07213mg of DDT in the extract / 5.29g spinach = <em>0.0136mg DDT / g spinach</em>

<em />

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3 years ago
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Los dos átomos operan simultáneamente
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3 years ago
What mass of carbon dioxide is produced from the complete combustion of 6.00×10−3 g of methane?
Crazy boy [7]

The answer for the following problem is explained below.

  • <u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

Explanation:

Given:

mass of methane = 6.00 × 10^-3 grams

CH_{4} + O_{2} → CO_{2} + H_{2}O

Firstly balance the following equation:

Before balancing the equation:

CH_{4} + O_{2} → CO_{2}  + H_{2} O

After balancing the equation:

CH_{4}  + 2O_{2} → CO_{2} + 2 H_{2} O

where;

CH_{4}  represents methane molecule

O_{2} represents oxygen molecule

CO_{2}  represents carbon dioxide molecule

H_{2}O  represents water molecule

 

       CH_{4} +2 O_{2} → CO_{2} + 2H_{2}O

        16  grams of methane       →        44 grams of carbon dioxide

        6 × 10^-3 grams of methane →          ?

                 = \frac{44*6.00*10^{-3} }{16}

          = 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.

<u><em>Therefore 16.5 × 10^-3 grams of carbon dioxide is produced from the complete combustion.</em></u>

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