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Rudik [331]
3 years ago
6

Consider the equation: 2NO2= N2O4. Using ONLY the information given by the equation which of the following changes would increas

e the molar concentration at equilibrium of the product N2O4?
Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
By the way I love science. So there are two moles of NO2(g) with an N204(g). Decreasing most moles. So your answer would be "Pressure Increase" Hope his helps.
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Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
nignag [31]

Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

  • In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:

N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

3 0
3 years ago
What is isostasy?
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Explanation:

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7 0
2 years ago
The average person in the United States is exposed to the following amount of radiations annually. Rank the following source of
Papessa [141]
<span>#1 is air radon, #2 is x-ray, #3 is ground, #4 is cosmic radiation, #5 is TV tube, #6 is weapons test fallout . That's all I got hope I helped!</span>
8 0
3 years ago
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Which of the following acids should be used to prepare a buffer with a pH of 4.5?A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4(COOH)2,
ioda

Answer:

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3 ) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

  • pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]

⇒ 1.8 E-5 = [H3O+](0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log ( 3.1623 E-5 ) = 4.5

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3 years ago
Rutherford's experiment showed that subatomic particles
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B. Don't take up most of the space in an atom,
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