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2 years ago

what is the general formula for straight-chain hydrocarbons with n carbon atoms and m (multiple) double bonds?

Chemistry

1 answer:

Katarina [22]2 years ago

8 0

the formula:

$C_{a} H_{2a+2-2x}$

where

a - number of the carbon atoms

x - number of double bonds

We take as a example the hydrocarbon with 4 carbon atoms

x = 1 → $C_{4} H_{2*4+2-2*1}$ → $C_{4} H_{8}$

x = 2 → $C_{4} H_{2*4+2-2*2}$ → $C_{4} H_{6}$

x = 3 → $C_{4} H_{2*4+2-2*3}$ → $C_{4} H_{4}$

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Given the following reactions Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ΔH = -28.0 kJ 3Fe (s) + 4CO2(s) → 4CO (g) + Fe3O4(s) ΔH =

Taya2010 [7]

Answer: The enthalpy of the reaction is -109 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Fe_2O_3(s)+3CO(s)\rightarrow 2Fe(s)+3CO_2(g)$ $\Delta H=-28.0kJ$ $\times 3$

$3Fe_2O_3(s)+9CO(s)\rightarrow 6Fe(s)+9CO_2(g)$ $\Delta H=-84.0kJ$ (1)

$3Fe(s)+4CO_2(s)\rightarrow 4CO(g)+Fe_3O_4(s)$ $\Delta H=+12.5kJ$ $\times 2$

$6Fe(s)+8CO_2(s)\rightarrow 8CO(g)+2Fe_3O_4(s)$ $\Delta H=+25.0kJ$ (2)

The final reaction is:

Subtracting (2) from (1):

$3Fe_2O_3(s)+CO(g)\rightarrow CO_2(g)+2Fe_3O_4(s)$ $\Delta H=-84.0-(+25.0)=-109kJ$

Thus the enthalpy of the reaction is -109 kJ

7 0

2 years ago

In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student genera

Paraphin [41]

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3

3 0

2 years ago

How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu

Alexxandr [17]

The moles of chromium (iii) nitrate produced is calculated as follows

write the equation for reaction

3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb

by use of mole ratio between Pb(NO3)2 to Cr(NO3)3 which is 3 : 2 the moles of Cr(NO3)3 is therefore
= 0.85 x2 /3 = 0.57 moles

5 0

2 years ago

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

The final concentration of hydroxide ion = 0.28 M

5 0

2 years ago

Read 2 more answers

Groups of atoms that are added to carbon backbones and give them unique properties are known as

Irina-Kira [14]

Answer:

Groups of atoms that are added to carbon backbones and give them unique properties are known as Functional Groups.

Explanation:

In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.

Examples:

S No. Functional Group Name

1 R--X Alkyl Halides

2 R--OH Alcohols

3 R--NH₂ Amines

4 R--O--R Ethers

5 R--CO--R Ketones

6 R--CO--H Aldehydes

7 R--CO--OH Carboxylic acids

8 R--CO--X Acid Halides

10 R--CO--NR₂ Acid Amides

11 R--CO-OR' Esters

3 0

2 years ago