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zlopas [31]
2 years ago
6

1.00 mole of O2 contains the same number of oxygen atoms as?

Chemistry
1 answer:
Sergio [31]2 years ago
5 0

Answer:

1.00mole of CH3CO2H

Explanation:

1mole of O2mole =26.022*10^23 atoms

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MatroZZZ [7]

Answer:

The molecules inside the ice pack are frozen therefore they are stuck in place. The molecules inside an insulated cup aren't frozen but still stuck inside of their container so that they can not fly out.

Explanation:

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3 years ago
The set of chemical and physical properties of a ______ are different from those of its constituents.
Veronika [31]

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compound

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hope this helps

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HELP WITH CHEMISTRY PLEASE!
maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

  • In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
  • Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

<u><em>In our problem:</em></u>

P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.


2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.


3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.


4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.


5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

<u><em>In our problem:</em></u>

P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

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3 years ago
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