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2 years ago

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What mass of propane (C3H8) is needed to produce 346 g carbon dioxide in the following reaction? A. 346 g C3H8 B. 1.86 g C3H8 C.

5075 g C3H8 D. 115 g C3H8

2 answers:

9966 [12]2 years ago

7 0

The answer to this question 116 g.

Kaylis [27]2 years ago

7 0

Answer : The correct option is, (D) 115 grams

Explanation :

Mass of $CO_2$ = 346 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CO_2$ .

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{346g}{44g/mole}=7.86moles$

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

The balanced chemical reaction is,

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

From the balanced reaction we conclude that,

As, 3 mole of $CO_2$ obtained from 1 mole of $C_3H_8$

So, 7.86 moles of $CO_2$ react to give $\frac{1}{3}\times 7.86=2.62$ moles of $C_3H_8$

Now we have to calculate the mass of $C_3H_8$ .

$\text{Mass of }C_3H_8=\text{Moles of }C_3H_8\times \text{Molar mass of }C_3H_8$

$\text{Mass of }C_3H_8=(2.62mole)\times (44g/mole)=115.28g=115g$

Therefore, the mass of propane needed are, 115 grams

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kiruha [24]

Explanation:

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2 years ago

Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r

denis23 [38]

Answer :

(1) The frequency of photon is, $3\times 10^{10}Hz$

(2) The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = $1.0cm=0.01m$ (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

$\lambda$ = wavelength of photon

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$\nu=\frac{3\times 10^8m/s}{0.01m}$

$\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz$ $(1Hz=1s^{-1})$

The frequency of photon is, $3\times 10^{10}Hz$

(2) Now we have to calculate the energy of photon.

Formula used :

$E=h\times \nu$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})$

$E=1.988\times 10^{-23}J/photon$

The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) Now we have to calculate the energy in J/mol.

$E=1.988\times 10^{-23}J/photon$

$E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)$

$E=11.97J/mol$

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

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3 years ago

Read 2 more answers

An atom of nitrogen has seven electrons. I hitch diagram correctly represents the arrangement of these electrons?

Alex787 [66]

Answer:

C.

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∧ ∧ ∧

⇅ ⇅ | | |

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3 years ago

Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(

Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

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3 years ago

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Gre4nikov [31]

Answer:

6.022⋅1023

Explanation:

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2 years ago