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2 years ago

Look back at parts A and B to compare the properties of the unknown elements with the properties of the known

Chemistry

1 answer:

valentina_108 [34]2 years ago

3 0

2 because it was a scam and

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CH4 (g) yields C(g) + 4H (g) (reaction for expansion)

anyanavicka [17]

Answer:

Here's what I find.

Explanation:

$\rm CH$_{4}$(g) $\, \rightleftharpoons \,$ C(g) + 4H(g)$

$\text{Same} =\begin{cases}1. & \text{The number of atoms of each element}\\2. & \text{The state of the methane molecules}\\4. &\text{The state of the carbon atoms}\\\end{cases}$

$\text{Different} =\begin{cases}3. & \text{The enthalpy change of the reaction}\\\end{cases}$

The sign of ΔH changes when you reverse the reaction.

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3 years ago

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2 years ago

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2 years ago

In a gas grill, 29 lbs propane C3H8 are

dimulka [17.4K]

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Given :

Mass of $C_3H_8$ = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of $C_3H_8$ = 44 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$ .

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_3H_8$ react to give 3 moles of $CO_2$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 3=896.7$ moles of $CO_2$

and,

As, 1 mole of $C_3H_8$ react to give 4 moles of $H_2O$

So, 298.9 mole of $C_3H_8$ react to give $298.9\times 4=1195.6$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$ .

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2$

$\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs$

and,

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O$

$\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs$

The total mass of products = Mass of $CO_2$ + Mass of $H_2O$

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.

6 0

3 years ago