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kkurt [141]
3 years ago
5

Identify the substance which has same number of molecules in 14g N2

Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
We need to know the number of molecules which is number of moles * Avogadro’s number
So
We need to compare number of moles because Avogadro’s numbers is constant for all
We’ll start with Nitrogen gas : 14g/28g.mol=0.5 mole

For NH3 : 3.4g / (14+3)g.mol= 0.2 mol
For NO. : 15g/(14+16)g.mol = 0.5 mol
For O2 : 64g/32 g.mol = 2 mol
For SO2: 32g /( 32+32) g.mol = 0.5mol
For H2. : 1g/ 2g.mol = 0.5 mol

So
H2& SO2 & NO have the same number of moles of N2 and the same number of molecules of N2
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Answer:

1) 6.0 atm.

2) 2.066 atm.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas.

V is the volume of the container.

n is the no. of moles of the gas.

R is the general gas constant.

T is the temperature of the gas (K).

<em>1) What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 2.0 atm and the temperature is held constant?</em>

  • At constant T and at two different (P, and V):

<em>P₁V₁ = P₂V₂.</em>

P₁ = 2.0 atm, V₁ = 150.0 mL.

P₂ = ??? atm, V₂ = 50.0 mL.

<em>∴ P₂ = P₁V₁/V₂</em> = (2.0 atm)(150.0 mL)/(50.0 mL) = <em>6.0 atm.</em>

<em>2. A sample of a gas in a rigid container at 30.0°C and 2.00 atm has its temperature increased to 40.0°C. What will be the new pressure?</em>

<em></em>

  • Since the container is rigid, so it has constant V.
  • At constant V and at two different (P, and T):

<em>P₁/T₁ = P₂/T₂.</em>

P₁ = 2.0 atm, T₁ = 30.0°C + 273 = 303 K.

P₂ = ??? atm, T₂ = 40.0°C + 273 = 313 K.

<em>∴ P₂ = P₁T₂/T₁ </em>= (2.0 atm)(313.0 K)/(303.0 K) =<em> 2.066 atm.</em>

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So if we continue to push at the truck and you feel it starting to move, then once it starts moving it is much easier to push, that is because we moved past static friction to rolling friction. Rolling friction is what helps slow things down. If you roll a ball across a carpet floor it eventually comes to a stop.

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In a study of the formation of NOx air pollution, a chamber heated to 2200°C was filled with air (0.790 atm N₂, 0.210 atm O₂). W
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Answer:

N₂ = 0.7515atm

O₂ = 0.1715atm

NO = 0.0770atm

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO(g)

Where Kp is defined as:

Kp = \frac{P_{NO}^2}{P_{N_2}P_{O_2}}}

Pressures in equilibrium are:

N₂ = 0.790atm - X

O₂ = 0.210atm - X

NO = 2X

Replacing in Kp:

0.0460 = [2X]² / [0.790atm - X] [0.210atm - X]

0.0460 = 4X² / 0.1659 - X + X²

0.0460X² - 0.0460X + 7.6314x10⁻³ = 4X²

-3.954X² - 0.0460X + 7.6314x10⁻³ = 0

Solving for X:

X = - 0.050 → False answer. There is no negative concentrations.

X = <em>0.0385 atm</em> → Right answer.

Replacing for pressures in equilibrium:

N₂ = 0.790atm - X = <em>0.7515atm</em>

O₂ = 0.210atm - X = <em>0.1715atm</em>

NO = 2X = <em>0.0770atm</em>

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