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3 years ago

Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:

Chemistry

1 answer:

hammer [34]3 years ago

5 0

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

<em>Where [] are equilibrium concentrations of each specie</em>

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As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

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A gas occupies 3.5L at 2.5 atm of pressure. What is the volume at 787 torr at the same temperature?

Kitty [74]

Answer: 8.45 L

Explanation:

Given that,

Initial volume (V1) = 3.5L

Initial pressure (P1) = 2.5 atm

[Since final pressure is given in torr, convert 2.5 atm to torr

If 1 atm = 760 torr

2.5 atm = 2.5 x 760 = 1900 torr

Final volume (V2) = ?

Final pressure (P2) = 787 torr

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

3.5L x 1900 torr = 787 torr x V2

6650L•torr = 787 torr•V2

Divide both sides by 787 torr

6650L•torr/787 torr = 787 torr•V2/787 torr

8.45 L = V2

Thus, the volume of the gas at 787 torr and at the same temperature is 8.45 Liters

5 0

3 years ago

Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?

pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0

3 years ago

Which of the following represents the least number of molecues?

Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

$molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}$

So, let´s get the number of molecules for each of the options.

$a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules$

$b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules$

$c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules$

$d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules$

$d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules$

the smalest number is in option a)

Best of luck.

7 0

2 years ago

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Flura [38]

Answer:

Explanation:

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8 0

2 years ago

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Mars2501 [29]

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3 years ago