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Tema [17]
2 years ago
6

What is the molecule shown below?

Chemistry
1 answer:
FrozenT [24]2 years ago
8 0

Answer:

2,3–dimethylpentane

Explanation:

To know which option is correct, we shall determine the name of the compound.

To obtain the name of the compound, do the following:

1. Determine the longest continuous carbon chain. This gives the parent name of the compound.

2. Identify the substituent group attached to the compound.

3. Locate the position of the substituent group by giving it the lowest possible count.

4. Combine the above to obtain the name of the compound.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 5. Thus, the parent name of the compound is pentane.

2. The substituent group attached is methyl (–CH₃)

3. There are two methyl group attached to the compound. One is located at carbon 2 and the other at carbon 3.

4. Therefore, the name of the compound is:

2,3–dimethylpentane

None of the options are correct.

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What's the question?

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A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

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Balance <br><br> _Mg + _HCL = _MgCl2 + H2
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_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side: 

_Mg + _HCl = _MgCl2 + H2

Mg- 1                 Mg-1
               
H-1                    H-2

Cl-1                    Cl-2

Mg is balanced on both sides so move on to the next (put a 1 in the space). 

1Mg

There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.

2HCl

Balamced equation will be: 

 1Mg + 2HCL = 1MgCl2 + H2
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Answer:

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c = 96

d = 86

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What comes in contact with Sodium Hydroxide to form soap and glycerine it is a chemical change because you cannot reverse this c
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Answer:

Your answer is triglycerides. Hope this helps.

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