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3 years ago

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For this question use the main assumptions of the Kinetic Molecular Theory of gases: 1. Gases are made up of molecules which are

relatively far apart. 2. The molecules are in motion at high speeds. 3. The molecular collisions are perfectly elastic. 4. Increase in temperature increases the kinetic energy of the molecules. The idea that no energy is lost when gas molecules hit the walls of a container or each other is explained by 1 2 3 4

1 answer:

Vera_Pavlovna [14]3 years ago

4 0

Answer:

Number 3

Explanation:

No energy is lost because collisions are perfectly elastic. Therefore, the answer is number 3.

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 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.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

Novay_Z [31]

<h3>Answer:</h3>

0.64 Moles of Propane

<h3>Explanation:</h3>

Data:

Moles of Carbon = 1.5 mol

Conversion factor = 7 mol C produces = 3 mol of Propane

Solution:

As we know,

7 moles of Carbon produces = 3 moles of Propane

Then,

1.5 moles of Carbon will produce = X moles of Propane

Solving for X,

X = (1.5 moles × 3 moles) ÷ 7 moles

X = 0.6428571 moles of Propane

Or rounded to two significant figures,

X = 0.64 Moles of Propane

5 0

2 years ago

Are the atoms really "sharing" electrons? Explain.

Iteru [2.4K]

Answer:

yes, in certain cases

there are different types of bondings between atoms

and in some they lend electrons to make their atom stable this type of bonding is called ionic bonding

and in covalent bond the atoms share their electrons

7 0

3 years ago

Seawater is around a 3% aqueous salt solution.

Eduardwww [97]

Answer:

the answer is distillation

3 0

1 year ago

Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M

liberstina [14]

Answer:

$= \mathbf{C_3H_6O}$

Explanation:

From the given information, since the molecular mass of the ion M+ is not given;

Let's assume M+ = 58.0423

So, by applying the 13th rule;

we will need to divide the mass by 13, after dividing it;

The quotient n = no. of carbon; &

The addition of the quotient (n) with the remainder r = no. of hydrogen.

So;

$\dfrac{58}{13}= 4 \ remainder \ 6$

So;

$C_nH_{n+r} = C_4H_{4+6}$

$= C_4H_{10}$

From the given information; we have oxygen present, so since the mass of oxygen = 16, we put oxygen in the molecular formula by removing $CH_4$ . Also, since the mass is an even number then Nitrogen is 0.

So, we have:

$= \mathbf{C_3H_6O}$

6 0

2 years ago