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alekssr [168]
3 years ago
11

What is the electron configuration for bromine

Chemistry
1 answer:
Alika [10]3 years ago
8 0

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5

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A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
Formula for Sodium Hydrogen Carbonate NaHCO3? or Na2HCO3?
yan [13]

NaHCO3 is the right answer

8 0
3 years ago
Which of the following best describes how an ionic compound dissolves in water?
Nadusha1986 [10]

The statement that best describes how an ionic compound dissolves in water is as follows: it separates into individual molecules and is an electrolyte, which is option C.

<h3>What is an ionic compound?</h3>

Ionic compound is any compound is a chemical compound composed of ions (charged atoms) held together by electrostatic forces termed ionic bonding.

Ionic compounds are electrolytes i.e. a substance when, in solution or when molten, ionizes and conducts electricity.

For example; sodium chloride (NaCl) is an ionic compound breaks down into sodium ions (Na+) and chloride ion (Cl-).

Therefore, the statement that best describes how an ionic compound dissolves in water is as follows: it separates into individual molecules and is an electrolyte.

Learn more about ionic compound at: brainly.com/question/9167977

#SPJ1

4 0
2 years ago
Calculate the molarity of 29.0g of ethanol (C2H5OH) in 545 mL of solution.
BigorU [14]
Molar mass ethanol:

C2H5OH = 12 x 2+ 1 x 5 + 16  + 1 =  46.0 g/mol

volume = 545 mL in liters: 545 / 1000 => 0.545 L

number of moles:

29.0 / 46.0 => 0.6304 moles

M = n / V

M = 0.6304 / 0.545

M = 1.156 mol/L

hope this helps!


5 0
3 years ago
Read 2 more answers
If the equation: FeCl3+O2--&gt;Fe2O3+Cl2, how many moles of chlorine gas can be produced if 4 moles of FeCl3 react with 4 moles
BlackZzzverrR [31]

Answer:

6 moles of Cl2

Explanation:

First, the equation has to be balanced, which makes it 4 FeCl3 + 3 O2 --> 2 Fe2O3 + 6 Cl2

Using this information, we can see that one mole of O2 will not be present in the reaction. Since four moles of FeCl3 are needed to react in the equation, which would produce six moles of Cl2, and only four moles of FeCl3 are present, six moles of Cl2 would be produced.

4 0
2 years ago
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