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3 years ago

What occurs during an exothermic change

2 answers:

6 0

In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.

nikklg [1K]3 years ago

3 0

The system releases energy to its surroundings.

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A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

7 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

12000 inches to yards

Nutka1998 [239]

ANSWER

$\begin{equation*} 333.33\text{ yds} \end{equation*}$

EXPLANATION

We want to convert 12000 inches to yards.

To do this, divide the value in inches by 36:

$\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}$

That is the answer.

4 0

1 year ago

What angle is formed by the sun, the earth, and the moon during an eclipse?.

Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

4 0

3 years ago

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and

taurus [48]

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is $V_1$ and initial pressure $P_1=1atm$ ( As pressure is constant )