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AURORKA [14]
3 years ago
10

Write the overall balanced equation for the reaction: N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)

Chemistry
1 answer:
zzz [600]3 years ago
8 0

Answer:

This reaction has infinite ways to be balanced

Explanation:

To balance this equation we can use the algebraic method:

N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)

Where we write each molecule as a letter:

A + B → C + D + E

Then, we write the equations according the number of atoms of each molecule. That is:

Oxygen → 4A + B = C + 2D + 2E <em>(1)</em>

Nitrogen → 2A = C + E <em>(2)</em>

Carbon → B = D <em>(3)</em>

Then, we have to give 1 arbitral number for a letter. For example:

B = 1; D = 1

<em>(1) </em>4A + 1 = C + 2 + 2E

4A = C + 2E + 1

2A = C + E <em>(2) </em>Twice <em>(2):</em>

4A = 2C + 2E

Subtracting <em>(1) </em>in <em>(2)</em>

C + 2E + 1 = 2C + 2E

C + 1 = 2C

1 = C

Si 1 = C:

4A + 1 = 1 + 2 + 2E

4A = 2 + 2E <em>(1)</em>

y:

2A = 1 + E <em>(2)</em>

Twice:

4A = 2 + 2E

As <em>(1) </em>and <em>(2) </em>are the same equation:

<h3>This reaction has infinite ways to be balanced</h3><h3 />

For example:

N2O4(g) + CO → NO(g) + CO2(g)+NO2(g)

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Which of the following can you physically change for a pure substance such as water, without altering the chemical composition?
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2) If I have an unknown quantity (# of moles) of gas at a pressure of 32 atm, a volume of 70 L, and a temperature of 25˚C. How m
astra-53 [7]

Answer:

91.5 mol

Explanation:

Volume of gas = 70 L

Temperature = 25°C

Pressure = 32 atm

Moles of gas  = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

25+273.15 = 298.15 K

By putting values,

32 atm × 70 L = n ×0.0821 atm.L /mol.K × 298.15 K

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n = 2240 atm.L / 24.48  atm.L /mol  

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3 0
3 years ago
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as
nikitadnepr [17]

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

Moles of copper(II) nitrate = \frac{469 g}{187.5 g/mol}=2.5013 mol

Moles of sodium sulfide = \frac{156 g}{78 g/mol}=2 mol

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

\frac{1}{1}\times 2mol= 2 mol of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

\frac{2}{1}\times 2mol=4 mol sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

7 0
3 years ago
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