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Anastasy [175]
3 years ago
5

What is the HONC 1234 rule

Chemistry
1 answer:
kirza4 [7]3 years ago
8 0
The HONC 1234 rule is a way to remember the bonding tendencies of hydrogen, oxygen, nitrogen, and carbon atoms in molecules. Hydrogen tends to form one bond, oxygen two, nitrogen three and carbon four.
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When two molecules of methanol (CH3OH) react with oxygen, they combine with three O2 molecules to form two CO2 molecules and fou
Alex17521 [72]

Answer:

188

Explanation:

For every 2 molecules of methanol reacted, 4 molecules of water are formed.  Use this relationship to solve.

2/4 = 94/x

2x = 376

x = 188

188 molecules of water will be formed.

3 0
3 years ago
Write a balanced nuclear equation for the β emission of the following isotopes
andrezito [222]

The balanced nuclear equation for the β emission of the following isotopes is seen below:

92                 92                0

Sr            ⇒ Y         +   e

38                  39             -1

<h3>What is Beta emission?</h3>

This is also known as beta decay in which a beta ray is emitted from an atomic nucleus.

The element formed during the beta emission of strontium is referred to as Yttrium.

Read more about Beta emission here brainly.com/question/16334873

#SPJ1

6 0
1 year ago
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
Mass is measured against a standard by using a balance. <br> a. True<br> b. False
jok3333 [9.3K]
Mass is measured against a standard by using a balance. The statement given above is a fact. Therefore the answer among the choices is letter "A" or "A. True". I hope this helps you on your assignment. 
6 0
3 years ago
Read 2 more answers
The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi
Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the <em>length in millimetres</em>

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert <em>millimetres to metres </em>

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

6 0
2 years ago
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