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2 years ago

If 5 L of butane is reacted what volume of carbon dioxide is produced ILL GIVE BRAINLIEST

Chemistry

1 answer:

Len [333]2 years ago

8 0

Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.

This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.

Next, let's determine how many moles of butane are burned. This is obtained by

5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.

Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.

We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).

V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.

Explanation:

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Why does mecury hardly evaporate in room temperature

Crazy boy [7]

Because elemental mercury is a liquid at a room temperature

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2 years ago

A chemistry instructor provides each student with 8 test tubes at the beginning of the year. If there are 28 students per class,

klemol [59]

Three classes: 28×3=84 students
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7 0

3 years ago

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

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

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

3 years ago

How many atoms are in 20.6 moles of curium

Harman [31]

Answer: 1.24 × 10^25

Explanation:

$\frac{20.6 moles}{1}$ × $\frac{6.022 * 10^{24} }{1 Mole}$

Using our knowledge in unit conversions, we know the mole units cancel each other out and all there's left is the atom unit. From here we can multiply the fractions and eventually we end with the number 124.0532 × 10^23

According to the scientific notation rules, the number to the left of the decimal cannot exceed 10 so we have to move the decimal to the left two spaces. With this change, we also have to change the exponent of the 10. Because we moved the decimal point two spaces to the left, that means we have 10^25.

4 0

3 years ago

The scientist at the weather office predicted the temperatures in the city route route for the next week I will be the weather t

Cloud [144]

Answer:

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Explanation:

6 0

3 years ago