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3 years ago

Which has a slightly greater mass in a nuclear reaction, the reactants or the products? why?

1 answer:

6 0

<span>The reactants have a slightly greater mass. In a nuclear reaction, a small amount of mass is converted to energy according to the equation E = mc2. The difference in mass is referred to as the mass defect.</span>

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Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

8 0

3 years ago

the average speeds of gas molecules in.cylinders A,b,c,d are 0.01 m/s, 0.005m/s ,0.1 m/s and 0.5 m/s, respectively .which cylind

Darya [45]

<span>The average speed of the gas is related to the kinetic energy of the gas. The kinetic energy of the gas is also related to the temperature of the gas. If the average speed of the gas is closer to zero, it means that it has very low motion or kinetic energy. This can be inferred that the gas has a very low temperature. At absolute zero, the motion of all the gas molecules stops. This means that the kinetic energy of the gas is also zero. Zero kinetic energy means zero average speed.</span> <span>So, the answer is cylinder B. The average speed of the gas in cylinder B is closest to zero.</span>

8 0

3 years ago

Як визначити амфотерний, луг чи основний елемент в хімії. Коли назви даєш, то пишеш наприклад CuSO4 - амф, купрум сульфат. Чому

Leno4ka [110]

What? Please write in Spanish or English

7 0

3 years ago

A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o

yaroslaw [1]

<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

Mass of the unknown metal sample as 58.932 g
Initial temperature of the metal sample as 101°C
Final temperature of metal is 23.68 °C
Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g
Initial temperature of water = 21°C
Final temperature of water is 23.68 °C
Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

= 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q = m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

= 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>

We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.

Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0

3 years ago

Write formula of compound given two chemicals

Evgesh-ka [11]

Nitrous oxide has 2 nitrogen atoms and 1 oxygen atom per molecule

3 0

3 years ago