Ask question

Ask question

All categories

3 years ago

6

Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon

dioxide. The molar mass of octane is 114.33 g/mole.
2C8H18 + 25O2 → 16 CO2 + 18 H2O

ΔH°rxn = -11018 kJ

1 answer:

natali 33 [55]3 years ago

8 0

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= $\frac{100.0 \ g}{114.33\ g/mole}$

= $0.8747 \ moles$

We know that 16 moles of $CO_2$ formation associates with -11018 kJ of heat, then

0.8747 moles $CO_2$ formation associates with,

= $-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat$

= $-0.0547\times 11018$

= $-602.69 \ KJ \ heat$

You might be interested in

How is oxygen different from neon

gtnhenbr [62]

Neon is an element and elements are pure oxygen is. Or it is a gas

8 0

3 years ago

Read 2 more answers

Which of the following statements correctly characterizes the subatomic particle

vichka [17]

Answer:

Its mass is about the same as that of a proton

Explanation:

4 0

3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

b)

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

What are the observable properties of salt and rocks, and explain how you used empirical evidence to explain that mixtures of so

777dan777 [17]

Answer:

Appearance. Pure rock salt is colorless. However, when found underground it is generally not completely pure, so may have yellow, red, gray or brown hues. It is either transparent or translucent and when you shine a light on it, its luster is vitreous, meaning it appears shiny and glassy.

Explanation:

6 0

2 years ago

Indirect costs incurred in a manufacturing environment that cannot be traced directly to a product are treated as a.product cost

Alisiya [41]

Answer:

C. product costs and expensed when the goods are sold

Explanation:

3 0

3 years ago