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2 years ago

A teacher told a learner to react benzene (CH) with chlorine (Cl₂) to

Physics

1 answer:

kolezko [41]2 years ago

3 0

The minimum quantity of benzene, C₆H₆ needed for the reaction is 106.67 g

<h3>How to determine the theoretical yield of chlorobenzene, C₆H₅Cl</h3>

From the question given, the following data were obtained

Actual yield = 100 g
Percentage yield = 65%
Theoretical yield =?

Percentage yield = (Actual / Theoretical) × 100

65% = 100 / Theoretical

0.65 = Actual / Theoretical

Cross multiply

0.65 × Theoretical = 100

Divide both sides by 0.65

Theoretical = 100 / 0.65

Theoretical yield = 153.85 g

<h3>How to determine the mass of benzene, C₆H₆ needed</h3>

Balanced equation

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Molar mass of C₆H₆ = 78 g/mol

Mass of C₆H₆ from the balanced equation = 1 × 78 = 78 g

Molar mass of C₆H₅Cl = 112.5 g

Mass of C₆H₅Cl from the balanced equation = 1 × 112.5 = 112.5 g

SUMMARY

From the balanced equation above,

112.5 g of C₆H₅Cl were obtained from 78 g of C₆H₆

Therefore,

153.85 g of C₆H₅Cl will be produced from = (153.85 × 78) / 112.5 = 106.67 g of C₆H₆

Thus, the minimum amount of benzene, C₆H₆ needed for the reaction is 106.67 g

Learn more about stoichiometry:

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A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

Fantom [35]

Centripetal acceleration of car is given by formula

$a_c = \frac{v^2}{R}$

now plug in the values in this

$a_c = \frac{23^2}{210}$

$a_c = 2.52 m/s^2$

Part b)

At position A we have

x component of acceleration is given as

$a_x = -a_c cos23$

$a_x = -2.32 m/s^2$

Part c)

At position A we have

y component of acceleration is given as

$a_y = -a_c sin23$

$a_y = -0.98 m/s^2$

Part d)

At position B we have

x component of acceleration is given as

$a_x = -a_c cos53$

$a_x = -1.52 m/s^2$

Part e)

At position B we have

Y component of acceleration is given as

$a_y = a_c sin53$

$a_y = 2.01 m/s^2$

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4 years ago

1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

$I=MR^2$

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

$I=(m)(r)^2=mr^2$

2) For a solid sphere, the moment of inertia is

$I=\frac{2}{5}MR^2$

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

$I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2$

3) For a disk of negligible thickness, the moment of inertia is

$I=\frac{1}{2}MR^2$

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

$I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2$

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

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3 years ago

Two blocks with mass M and 3M on a horizontal frictionless surface are pushed together and compress a spring of negligible mass

Naily [24]

Answer:

launching speed of the lighter block = -6 m/s

Explanation:

We are given;

Mass of light block; M

Mass of heavy block; 3M

Speed of launched block: v = 2m/s

We are told that the two blocks are sitting on the horizontal frictionless surface. Thus, we can say that no external force is being applied on the system and so, the momentum of the whole system is conserved accordingly to that condition.

We are also told that when the rope is cut with scissors, that the heavier block attains the speed of 2 m/s in the positive x-direction which is horizontal direction.

We know that formula for momentum is; M = mass x velocity.

Thus, the momentum of the heavier block is calculated as;

M_1 = 3M × 2

M_1 = 6M kg.m/s

Since no external force is applied on the object, the initial momentum will be zero.

Hence, to conserve the system, the momentum of the lighter block will be equal and opposite to the momentum of heavier block.

So, momentum of lighter block is;

M_2 = -6M kg.m/s

Since mass of lighter block is M and formula for momentum = mass x velocity.

Thus;

-6M = Mv

Where v is speed of lighter block.

So, v = -6M/M

v = -6 m/s

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3 years ago

Use the mass and density data to calculate the volume of corn syrup to the nearest tenth.

Nataly [62]

41.5 is the answer that i got. hope this helps!

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3 years ago

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An automobile is sitting on a hill which is 20 m higher than ground level. Find the mass of the automobile if it contains 362,60

mr Goodwill [35]

M= ?
g=9.8 m/s (2)
h=20 m

Eg=362,600 J
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3 years ago