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3 years ago

Examine the reactants of the incomplete double displacement reaction.

Physics

1 answer:

Bumek [7]3 years ago

8 0

<h3>Answer:</h3>

AgCl + NaNO₃

<h3>Explanation:</h3>

The reaction between silver nitrate and sodium chloride is an example of a double displacement reaction.
In a double displacement reaction compounds or salts reacts and exchange cations or anions to form new compounds or salts.
In this case, silver nitrate and sodium chloride exchange anions and cations to form silver chloride and sodium nitrate.

Therefore, the complete reaction is given by;

AgNO3 + NaCl → AgCl + NaNO₃

But since silver chloride is a precipitate, the reaction may also be an example of a precipitation reaction.

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Study the image of the moving car.

lora16 [44]

Answer:

Low Potential energy and High Kinetic energy

Explanation:

Hope this helps and have a good day! Apologies if it's wrong.<3

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2 years ago

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Two facts about Saturns largest moon

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The diameter of its sold body is 5,150 km(3,200 miles). The moon is called Ganymede.

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3 years ago

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A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Oduvanchick [21]

Answer : The heat capacity of the calorimeter is, $6.72J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of water = $4.184J/g^oC$

$c_2$ = specific heat of calorimeter = ?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of mixture = $35.0^oC$

$T_1$ = initial temperature of water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now put all the given values in the above formula, we get

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Therefore, the heat capacity of the calorimeter is, $6.72J/g^oC$

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3 years ago

A racing car starts from rest at t=0 and reaches a final speed v at time t. If the acceleration of the car is constant during th

Readme [11.4K]

Answer:

All true statements are shown: a) <em>The average speed of the car is </em> $\frac{v}{2}$ , c)<em>The magnitude of the acceleration of the car is </em> $\frac{v}{t}$ .

Explanation:

Let prove the validity of each statement:

a) <em>The average speed of the car is </em> $\frac{v}{2}$ .

The average speed ( $\bar v$ ) is defined by the following formula:

$\bar v = \frac{v_{o}+v_{f}}{2}$ (1)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the racing car.

If we know that $v_{o} = 0$ and $v_{f} = v$ , then the average speed of the racing car:

$\bar v = \frac{0+v}{2}$

$\bar v = \frac{v}{2}$

The statement is true.

b) <em>The car travels a distance </em> $v\cdot t$ .

Since the racing car is accelerating uniformly, the distance travelled by the car is represented by the following kinematic formula:

$x - x_{o}=v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $a$ is the acceleration of the racing car, measured in meters per square second.

The statement is false.

c) <em>The magnitude of the acceleration of the car is </em> $\frac{v}{t}$ .

Since the racing car is accelerating uniformly, the velocity of the racing car is represented by the following kinematic formula:

$v_{f} = v_{o}+a\cdot t$ (3)

Then, we clear the acceleration of the expression:

$a = \frac{v_{f}-v_{o}}{t}$

If we know that $v_{o} = 0$ and $v_{f} = v$ , then the acceleration of the car is:

$a = \frac{v-0}{t}$

$a = \frac{v}{t}$

The statement is true.

d) <em>The velocity of the car remains constant</em>.

Since the car accelerates uniformly, the vehicle does not travel at constant velocity.

The statement is false.

8 0

3 years ago

A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a spe

Amiraneli [1.4K]

Answer:

$v_r=13.68\ m.s^{-1}$ is the final velocity of the racket.

$F=18.86\ N$

Explanation:

Given:

mass of the racket, $m_r=1000\ g$

mass of ball, $m_b=60\ g$

initial speed of racket, $u_r=15\ m.s^{-1}$

initial speed of ball, $u_b=18\ m.s^{-1}$

final speed of ball, $v_b=40\ m.s^{-1}$

time of contact of racket with the ball, $t=0.07\ s$

<u>By the law of conservation of momentum:</u>

$m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b$

where: $v_r=$ final velocity of the racket

$1000\times 15+60\times 18=1000\times v_r+60\times 40$

$v_r=13.68\ m.s^{-1}$ is the final velocity of the racket.

<u>By the Newton's second law of motion:</u>

$F=\frac{dp}{dt}$ ............................(1)

where:

dp = change in momentum

dt = change in time

Change in momentum of ball:

$\Delta p_b=m_b.v_b-m_b.u_b$

$\Delta p_b=60\times 10^{-3}\times (40-18)$

$\Delta p_b=1.32\ kg.m.s^{-1}$

Now, using eq.(1):

$F=\frac{1.32}{0.07}$

$F=18.86\ N$

3 0

4 years ago