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3 years ago

Calculate the heat energy released when 17.7 g of liquid mercury at 25.00 °c is converted to solid mercury at its melting point.

Physics

1 answer:

maria [59]3 years ago

8 0

Answer;

Q = 359.2-J

Explanation;

Given that;

Constants for mercury at 1 atm

Heat capacity of Hg(l) is 28.0 J/(mol*K)

melting point is 234.32 K

Enthalpy of fusion is 2.29 kJ/mol

17.7-g Hg / 200.6g/mol = 0.0882 mol Hg;

°C + 273 = 298 K;

2.29-kJ/mol = 2290-J/mol

Q = (m x ΔT x Cp) + (m x Hf)

Q = 0.0882-mol x (298 - 234.32) x 28.0-J/mol*K) + (0.0882-mol x 2290-J/mol)

Q = 157.26-J + 201.978-J

Q = 359.2-J

Q=359-J (3 sig fig allowed due to 17.7-g given in problem)

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At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

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How does earths magnetic field protect us from solar flares

Alex787 [66]

Answer:

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2 years ago

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a car takes off from rest and covers a distance of 80m on a straight road in 10s.calculate the magnitude of its acceleration

olasank [31]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Here's the solution :

Let's find the final velocity :

$\dfrac{displacement}{time}$

$\dfrac{80}{10}$

$8 \: \: ms {}^{ - 1}$

Initial velocity (u) = 0 (cuz it started from rest)

Final velocity (v) = 8 m/s

Time taken (t) = 10 sec

now, we know that :

$acceleration = \dfrac{v - u}{t}$

$\dfrac{8 - 0}{10}$

$\dfrac{8}{10}$

$0.8 \: \: m {s}^{ - 2}$

Acceleration = 0.8 m/s²

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3 years ago

Unit of work is derived unit why

adoni [48]

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Acceleration is the change in distance over time. true or false

olganol [36]

the answer is true acceleration is the change in distance over time.
hope this helps

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