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ludmilkaskok [199]
3 years ago
5

Determine the value of the equilibrium constant, Kgoal, for the reaction N2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=? by making use of

the following information: 1. N2(g)+O2(g)⇌2NO(g), K1 = 4.10×10−31 2. N2(g)+2H2(g)⇌N2H4(g), K2 = 7.40×10−26 3. 2H2O(g)⇌2H2(g)+O2(g), K3 = 1.06×10−10 Express your answer numerically.
Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

Answer:

K = 1.79x10⁻³³

Explanation:

Using Hess's law, it is possible to fin K of a reaction by the algebraic sum of another related reactions.

In the reactions:

1. N2(g)+O2(g) ⇌ 2NO(g), K1 = 4.10×10−31 2.

2. N2(g)+2H2(g) ⇌ N2H4(g), K2 = 7.40×10−26

3. 2H2O(g)⇌2H2(g)+O2(g), K3 = 1.06×10−10

The sum of 1/2 (1) + 1/2 (2) produce:

N2(g) + 1/2O2(g) + H2(g) ⇌ 1/2N2H4(g) + NO(g)

And K' = √4.1x10⁻³¹×√7.4x10⁻²⁶ = 1.74x10⁻²⁸

Now, this reaction + 1/2 (3):

N2(g) + H2O(g) ⇌ NO(g) + 1/2N2H4(g)

And K of reaction is:

1.74x10⁻²⁸×√1.06x10⁻¹⁰ = <em>1.79x10⁻³³</em>

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For 2,663 kg of a compound with the formula Al(SO), determine the following quantities (4 pts each); a) The number of moles of t
Nastasia [14]

Answer:

a) 35.485 moles of Al(SO)

b) 35.485 moles of S atoms

c) 2.136197(10^{25}) Al atoms

d) 567.723 g of O

Explanation:

Let's define the following terms :

1 mol = 6.02.(10^{23}) elemental units

For example :

1 mol of oxygen atoms = 6.02.(10^{23}) oxygen atoms

Now, our compound has the following formula

Al(SO)

Where Al is aluminium

S is sulfur

And O is oxygen

All the subscripts are 1 so we can say the following :

1 molecule of Al(SO) has 1 atom of Al , 1 atom of S and 1 atom of O

In terms of moles :

1 mol of Al(SO) has 1 mol of Al , 1 mol of S and 1 mol of O

The molar masses of Al, S and O are

molarmass_{(Al)}=26.982\frac{g}{mol}

molarmass_{(S)}=32.065 \frac{g}{mol}

molarmass_{(O)}=15.999\frac{g}{mol}

If we sum all the molar masses =(26.982+32.065+15.999)\frac{g}{mol}=75.046\frac{g}{mol}

Finally, 75.046 g of Al(S0) is 1 mol of Al(SO) which contains 26.982 g of Al, 32.065 g of S and 15.999 g of O.

1 mol of Al(SO) contains 1 mol of Al, 1 mol of S and 1 mol of O.

Now we can calculate a),b),c) and d)

For a)

2.663 kg=2663g

75.046 g of Al(SO) = 1 mol of Al(SO)

2663 g of Al(SO) = x

x=\frac{2663}{75.046}mol=35.485 mol

2.663 kg of Al(SO) contains 35.485 moles of Al(SO)

b) and c) 1 mol of Al(SO) molecules contains 1 mol of S atoms and 1 mol of Al atoms

We have 35.485 moles of Al(SO) molecules so

We have 35.485 moles of S atoms

And 35.485 moles of Al atoms

If 1 mol = 6.02(10^{23})

35.485 moles of Al have (35.485)(6.02)(10^{23})=2.136197(10^{25}) Al atoms

d) 75.046 g of Al(SO) contains 15.999 g of O

2663 g of Al(SO) contains x g of O

x=\frac{(2663).(15.999)}{75.046} g

x = 567.723 g of O

6 0
3 years ago
Atoms That have a positive charge will be attracted to atoms with a ___ charge.
san4es73 [151]

Answer:

Opposites attract, It will be attracted to a negative charge

Explanation:

3 0
3 years ago
2. 14g of Nitrogen gas and 8.0g of hydrogen react to produce ammonia according to the equation: N2 + 3H2 -- 2NH3 Calculate the m
GREYUIT [131]

Based on the equation of the reaction, nitrogen is the limiting reagent while hydrogen is the excess reagent.

<h3>What is the mole ratio of hydrogen to nitrogen in the formation of ammonia?</h3>

Hydrogen and nitrogen combines to form ammonia ina mole ratio of 3 : 1 as shown by the equation of the reaction below:

  • N2 + 3H2 -- 2NH3

The number of moles of the reactants in 14g of Nitrogen gas and 8.0g of hydrogen is calculated as follows:

  • Moles = mass/molar mass

Molar mass of N_{2} = 14.0 g

Molar mass of H_{2} = 2.0 g

Moles of N_{2} = 14/14.0 = 1 mole

Moles of H_{2} = 8/2.0 = 4 moles

Based on the data above:

  • The limiting reagent is nitrogen gas as it will be used up while hydrogen will be left over.
  • The moles of nitrogen is 1 mole
  • Hydrogen is the excess reagent and 1 mole will be left over
  • 3 moles of hydrogen will react with 1 mole of the nitrogen gas
  • mass of 3 moles of hydrogen is 3 × 2.0 g = 6.0g

Therefore, the limiting reagent is nitrogen while hydrogen is the excess reagent.

Learn more about limiting reagent at: brainly.com/question/24945784

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If the pressure of a sample of gas is doubled at constant temperature, what happens to the volume of the gas?
patriot [66]
The volume gas will stay constant 
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