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AURORKA [14]
2 years ago
5

7. Diethyl ether burns in air according to the following equation. C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l) If 7.15 L of CO2 is

produced at a temperature of 125°C and a pressure of 1.02 atm, what volume of oxygen, measured at STP, was consumed and what mass of diethyl ether was burned?
Chemistry
1 answer:
iren [92.7K]2 years ago
6 0

From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.

<h3>What is combustion?</h3>

Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)

We can obtain the number of moles of CO2 from;

PV = nRT

n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K

n = 7.29 /32.6

n = 0.22 moles

If 6 moles of oxygen produces 4 moles of CO2

x moles of oxygen produces 0.22 moles of CO2

x = 0.33  moles

1 mole of oxygen occupies 22.4 L

0.33 moles of oxygen occupies 0.33 moles *  22.4 L/ 1 mole

= 7.4 L of oxygen

Learn more about stoichiometry: brainly.com/question/13110055

#SPJ1

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1) A car is traveling down the interstate at 37.1 m/s. The driver sees a cop and quickly slows down. If the driver slows to 29.8
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1)

The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity of the car to change from u to v

In this problem, for this car we have:

u = 37.1 m/s

v = 29.8 m/s

t = 3 s

So, the acceleration is:

a=\frac{29.8-37.1}{3}=-2.43 m/s^2

2)

The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:

W=Fd

where

F is the force applied

d is the displacement of the box

Here we have:

F = 87.3 N is the force applied

d = 2.04 m is the displacement of the box

So, the work done to lift the box is:

W=(87.3)(2.04)=178.1 J

3)

The power is the rate of work done per unit time. It is calculated as:

P=\frac{W}{t}

where

W is the work done

t is the time taken to do the work

For the child in this problem, we have:

W = 1250 J is the work done by the child running up the stairs

P = 267 W is the power used

Therefore, re-arranging the equation, we find the time taken:

t=\frac{W}{P}=\frac{1250}{267}=4.68 s

4)

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the rabbit in this problem, we have:

m = 8.642 kg is the mass of the rabbit

KE = 125.6 is its kinetic energy

Solving the formula for v, we find the speed of the rabbit:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s

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The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by

\eta = \frac{E_{out}}{W_{in}}\cdot 100

where

E_{out} is the energy in output

W_{in} is the work in input

For the machine in this problem,

W_{in}=120 J is the work in input

E_{out}=93 J is the energy in output

Therefore, the efficiency of this machine is:

\eta=\frac{93}{120}\cdot 100=77.5\%

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During a collision, the total momentum of the system is always conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 =(m_1+m_2)v

where

m_1=212 kg is the mass of the first car

u_1=8.00 m/s is the initial velocity of the first car

m_2=196 kg is the mass of the 2nd car

u_2=6.75 m/s is the initial velocity of the 2nd car

v is the final velocity of the two cars stuck together (after the collision, they move together)

Solving the equation for v, we find:

v=\frac{m_1 u_1 +m_2 u_2}{m_1 +m_2}=\frac{(212)(8.00)+(196)(6.75)}{212+196}=7.40 m/s

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The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

v=f\lambda

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v is the speed of the wave

f is the frequency of the wave

\lambda is the wavelength

For the wave in the string in this problem we have:

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f = 12 Hz (frequency)

So, the speed of the wave is:

v=(12)(0.23)=2.76 m/s

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The relationship between frequency and wavelength for an electromagnetic wave is given by

c=f\lambda

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c is the speed of light in a vacuum

f is the frequency of the wave

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For the blue light in this problem, we have

f=6.2\cdot 10^{14}Hz (frequency)

while the speed of light is

c=3.0\cdot 10^8 m/s

So, the wavelength of blue light is:

\lambda=\frac{c}{f}=\frac{3.0\cdot 10^8}{6.2\cdot 10^{14}}=4.8\cdot 10^{-7} m

9)

The sound wave in this problem travels with uniform motion (=constant velocity), therefore we can use the following equation:

d=vt

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d is the distance covered by the wave

v is the speed of the wave

t is the time elapsed

In this problem:

v = 343 m/s is the speed of the sound wave

t = 0.287 s is the time elapsed

So, the distance covered by the wave is

d=(343)(0.287)=98.4 m

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