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sergey [27]
2 years ago
14

2 upper N upper O subscript 2 (g) right arrow 2 upper N upper o (g) plus upper O subscript 2 (g). Second: 2 upper N upper O (g)

right arrow upper N subscript 2 (g) plus upper O subscript 2 (g). Third: upper N subscript 2 (g) plus 2 upper O subscript 2 (g) right arrow upper N subscript 2 upper O subscript 4 (g). What is the equation for the overall reaction obtained by adding these equations?
Chemistry
1 answer:
Ipatiy [6.2K]2 years ago
5 0

The equation for the overall reaction obtained by adding these equations is;  2NO₂(g) ⇒ N₂O₄(g)

<h3>How to simplify chemical equations?</h3>

We are given equations;

2NO₂(g) ----> 2NO(g) + O₂(g)   ---(1)

2NO(g) -----> N₂(g) + O₂(g)   -----(2)

N₂(g) + 2O₂(g) ----> N₂O₄(g)   -----(3)

Add eq 1 and eq 2 to get;

2NO₂(g) + 2NO(g) ⇒ 2NO(g) + O₂(g) + N₂(g) + O₂(g)

2NO will cancel out to get;

2NO₂(g) ⇒ N₂(g) + 2O₂(g)   -----(4)

Add eq 3 to eq 4 to get;

N₂(g) + 2O₂(g) + 2NO₂(g) ⇒ N₂O₄(g) + N₂(g) + 2O₂(g)

This will reduce to;

2NO₂(g) ⇒ N₂O₄(g)

The correct question is;

Consider the chemical equations shown here.

2NO₂(g) ----> 2NO(g) + O₂(g)

2NO(g) -----> N₂(g) + O₂(g)

N₂(g) + 2O(g) ----> N₂O₄(g)

What is the equation for the overall reaction obtained by adding these equations?

Read more about chemical equations at; brainly.com/question/19837884

#SPJ1

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The number of sigma and pi bonds are,

          Sigma Bonds  =  16

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Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

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How many moles are in 20 grams of O₂ gas?
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Quantum transitions that result in the characteristic sharp lines of the X-ray spectrum always involve
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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
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