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FromTheMoon [43]
3 years ago
11

How does the energy of the activated complex compare with the energies of reactants and products? select one:

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
5 0

Answer:

  • Option d. i<u><em>t is higher than the energy of both reactants and products</em></u>

Explanation:

<em>Activated complex</em>, also known as transition state, is the intermediate structure formed in the course of a chemical reaction.

The activated complex is very unstable and of short life: it is at the peak of the potential chemical diagram, and can transform either into the reactants (backward) or the products (forward).

The activation energy of the reaction is the energy needed to reach the activated complex, then both reactants and products are lower in potential chemical energy than the activated complex, which is what explains why the activated complex can transform into one or another, reactants or products.

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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0
2 years ago
BRAINLIEST ASAP!!! Help
Andru [333]

Answer:

N2C14

Explanation:

<em> determined the bond type by looking if it is a metal or nometal</em>

<em>Ionic Bond:NM+M</em>

<em>Covalent Bond:NM+NM</em>

7 0
3 years ago
What is the volume of one mole of any gass at STP?
fenix001 [56]
Every mole is 22.4 L at STP
7 0
3 years ago
An insoluble solid is placed in water and the system allowed to reach equilibrium. The ratio of the rate at which ions join the
solmaris [256]

Answer:

One

Explanation:

Because it reaches an equibrum state so it's equals to one

7 0
3 years ago
What is the molar mass of (NH4)3 PO4? 113g, 121g, 149g, 339g
allochka39001 [22]
(NH4)3PO4 :

N = 14 * 3 =  42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
-------------------------
42+12+31+64 =  149 g / mol

Hope this helps!.


7 0
3 years ago
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