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3 years ago

Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa.

Chemistry

1 answer:

Schach [20]3 years ago

8 0

Answer:

31.2K

Explanation:

The following data were obtained from the question:

n = 1.42 mol

V = 3L

P = 123 kPa = 123000Pa

Recall: 101325Pa = 1atm

123000Pa = 123000/101325 = 1.21atm

R = 0.082atm.L/Kmol

T =?

Using the ideal gas equation PV = nRT, we can obtain the temperature as follows:

PV = nRT

T = PV/nR

T = (1.21 x 3)/(1.42 x 0.082)

T = 31.2K

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Which pair will for an ionic compound

vovikov84 [41]

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Explanation:

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Explanation:

The chart below shows monatomic ions formed when an atom loses or gains one or more electrons, and the ionic compounds they form. You can check your periodic table to see that the cations are monatomic ions formed from metals, and the anions are monatomic ions formed from nonmetals.

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2 years ago

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2 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

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= 6.8078 mg/l

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$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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3 years ago

A high pH value indicates _____.

Mumz [18]

Answer:

A high pH value indicates a high concentration of OH- ion

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3 years ago

What is the right answer for this. I really need help

statuscvo [17]

<span>así que te está diciendo que</span>

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3 years ago