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laiz [17]
3 years ago
12

When 5.0 moles of butane react completely?

Chemistry
1 answer:
Korolek [52]3 years ago
7 0

Moles of Carbondioxide-CO₂ produced = 20 moles

<h3>Further explanation</h3>

The combustion of hydrocarbons with excess oxygen will produce carbon dioxide and water(CO₂+H₂O), whereas if there is not much oxygen, carbon monoxide and water(CO+H₂O) will be obtained.

The reaction coefficient in a chemical equation shows the mole ratio of the reacting compounds

Reaction (combustion of butane) :

<em>2C₄H₁₀+13O₂⇒8CO₂+10H₂O</em>

Butane reacts completely, then Butane is the limiting reactant and oxygen as the excess reactant, so the moles of Carbon dioxide are based on the butane moles as the limiting reactant.

moles of butane - C₄H₁₀ = 5 moles

From the reaction, the mol ratio of C₄H₁₀ and CO₂ : 2 : 8, so mol CO₂ :

\tt \dfrac{8}{2}\times 5~moles=20~moles

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There will be three electrons in the outer shell.
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The pressure of a 250 mL sample of gas is 105 kPa. What would be the pressure if the volume were increased to 375 ml?
Arte-miy333 [17]

Answer:

<h2>The answer is 70 kPa</h2>

Explanation:

In order to find the pressure if the volume were increased to 375 ml we use Boyle's law

That's

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

From the question

P1 = 105 kPa = 105 , 000 Pa

V1 = 250 mL

V2 = 375 mL

So we have

P_2 =  \frac{105000 \times 250}{375}  =  \frac{26250000}{375}  \\  = 70000

We have the final answer as

<h3>70 kPa</h3>

Hope this helps you

8 0
4 years ago
Alpha particles are most like...
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4 years ago
What is the wavelength in picometers for an electron with 360 eV of kinetic energy?
Rom4ik [11]

Answer:

Wavelength = 64.635 pm

Explanation:

The expression for the deBroglie wavelength and kinetic energy is:

\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}

Where,  

\lambda is the deBroglie wavelength  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

m is the mass of electron having value 9.11\times 10^{-31}\ kg

K.E. is the kinetic energy of the electron.

Given, K.E. = 360 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, K.E. = 360\times 1.6022\times 10^{-19}\ J=5.76792\times 10^{-17}\ J

Applying in the equation as:

\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}

\lambda=\frac{6.626\times 10^{-34}}{\sqrt {{2\times 9.11\times 10^{-31}\times 5.76792\times 10^{-17}}}}\ m

\lambda=\frac{10^{-34}\times \:6.626}{\sqrt{10^{-48}\times \:105.0915024}}\ m

\lambda=6.4635\times 10^{-11}\ m

Also, 1 m = 10¹² pm

<u>So, Wavelength = 64.635 pm</u>

4 0
4 years ago
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