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GarryVolchara [31]
3 years ago
10

1 Starting with methane, name

Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

See Explanation and image attached

Explanation:

Methane is an alkane. The most common reaction for which alkanes are known is substitution reaction. In a substitution reaction involving alkanes, , one or more atoms of hydrogen is/are replaced.

In the presence of sunlight and molecular chlorine gas, homolytic fission of Cl2 takes place leading to the production of chlorine radicals.

These radicals react with methane in several propagation steps as shown  until the tetrachlorination product is finally obtained.

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ΑC2H6O+βO2 → γC2H4O2+δH2O
olga2289 [7]
Balancing the equation, we get:

2C₂H₆O + O₂ → 2C₂H₄O₂ + 2H₂O

So 
α = 2
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Under which conditions will sugar most likely dissolve fastest in a cup of water? open study
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This is how osmium appears in the periodic table.A purple box has O s at the center and 76 above. Below it says osmium and below
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Answer:

114

Explanation:

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8 0
3 years ago
Read 2 more answers
A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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4 years ago
Which of the following is not a response?only one
HACTEHA [7]

Answer:

'djweidj

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Explanation:

7 0
3 years ago
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