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RSB [31]
4 years ago
6

The hydrocarbon used to manufacture foam plastics is called styrene. Analysis shows composition to be 92.25% C and 7.75% H. Anal

ysis also shows a molar mass of 104g/mol. Determine the empirical and molecular formula for styrene.
Physics
1 answer:
zhannawk [14.2K]4 years ago
8 0

Answer:

molecular formula = C_{8} H_{8}

Explanation:

Given data

c = 92.25%

H = 7.75%

molar mass = 104 g/mol

to find out

the empirical and molecular formula for styrene

solution

we know that

styrene 1 g contain = 0.9225 g C and 0.0775 g H

so

C = 104 × 0.9225 g / 12 g/mol

C = 7.995 mol = approx 8 mol

and

H = 104 × 0.0775 g / 1 g/mol

H = 8.06 mol = approx 8 mol

so we say that 1 mole of styrene  have 8 mole of C and H

so

molecular formula = C_{8} H_{8}

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In a young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe i
dem82 [27]

Answer:

(d/λ) = 17.67

Explanation:

Equation for double-slit:

d*sinθ = m*λ

Rearrange the equation

(d/λ) = m/(sinθ)

Since it is at the second dark fringe, m = 2

Therefore:

(d/λ) = 2/(sin(6.5)

(d/λ) = 17.67

6 0
3 years ago
A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a
GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

7 0
3 years ago
A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field
Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0
3 years ago
A bicyclist with a mass of 50 kg is traveling at a rate of 30 m/s. It accelerates to a rate of 50 m/s in 5 seconds. What is the
balandron [24]

Answer:

F=m*(v^2/r)

F=82*(8^2/30)

F=174.9N

Explanation:

brainlest pls

8 0
3 years ago
Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet
yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2)  and the remainder leaves at (3)

The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

m_{3}=m_{1}-m_{2}

m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

We need to calculate the inlet velocity

Using formula of velocity

v=\dfrac{m}{\rho A}

Put the value into the formula

v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

v=21.9\ m/s

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0
4 years ago
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