Ask question

Ask question

All categories

3 years ago

13

A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.

21 kJ/°C. If the temperature rose from 25.0°C to 75.0°C, determine the value of ΔH°comb for substance.

1 answer:

aleksandrvk [35]3 years ago

4 0

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

$m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c}$ Where $m_{c}$ is the calorimeter mass and $n_{s}$ is the number of moles of the samples; $H_{c}$ is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product $m_{c} *Cp$ is the heat capacity, so the balance equation is:

$6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}$

So, the enthalpy of combustion can be calculated:

$H_{c}=-5899.5\frac{kJ}{mol}$

I will be happy to solve any doubt you have.

You might be interested in

What is the name of the protein did the human use to create dogs?

Margaret [11]

Humans did not create dogs.

Humans have never created any new living species.

So far, the only one who can do that is You Know Who.

3 0

3 years ago

Which event is an example of melting?

qaws [65]

Answer:

welding metal

Explanation:

5 0

2 years ago

Does uniform acceleration means increasing acceleration?

tatiyna

Yes. It means that the acceleration increases at a constant rate, for example 3 mph every second.

5 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers

Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

6 0

3 years ago