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3 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]3 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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The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

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3 years ago

A combustion reaction occurs between one molecule of methane (CHa) and two molecules of oxygen (O2).

SCORPION-xisa [38]

The total number of atoms make up the products :

D) 1 carbon, 4 hydrogen, and 4 oxygen

<h3>Further explanation </h3>

Complete combustion of Hydrocarbons with Oxygen will produce CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Hydrocarbon combustion reactions (especially alkanes)

For combustion of methane (CH₄) and two molecules of oxygen (O₂).

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

The number of atoms make up the products

CO₂ : 1 carbon, 2 oxygen

2H₂O : 4 hydrogen , 2 oxygen

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3 years ago

The rock in a particular iron ore deposit contains 78% Fe2O3 by mass. How many kilograms of the rock must a mining company proce

Studentka2010 [4]

Answer:

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Explanation:

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3 years ago

Although both N2 and 02 are naturally present in the air we breathe, high levels of NO and NO2 in the atmosphere occur mainly in

joja [24]

Answer:

(a) Increasing the temperature adds heat which is a reactant shifting the equilibrium rightwards.

(b) Pressure has no effect since the change in the number of moles is zero.

Explanation:

Hello,

In this case, one could represent the given reaction as:

$N_2(g)+O_2(g)+ \Delta _rH \leftrightarrow 2NO$

Since it is endothermic. Thus, solving the (a) statement, one identifies the heat as a reagent, that is why the reaction cools down as it progress, therefore, by increasing the temperature, heat is added, that is, a reagent is added, which shifts the equilibrium rightwards, in other words, more NO is produced so its concentration increases.

Furthermore, for the (b) statement, since the change in the number of moles is zero, based on the stoichiometric coefficients as shown below:

$\Delta \nu =2-1-1=0$