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Thepotemich [5.8K]
2 years ago
12

Zinc metal (Zn) and sulfur powder (s) undergo a chemical reaction to form zinc sulfide (ZnS) which equation represents this chem

ical reaction? S → ZnS + Zn ZnS → Zn + S Zn + S → ZnS Zn → ZnS + S
Physics
1 answer:
dexar [7]2 years ago
8 0

Zinc metal (Zn) reacts with sulfur (S) to create zinc sulfide (ZnS), and the chemical reaction is: Zn (s) + S (s) = ZnS (s).

Importance of Zinc (Zn) and sulfur (S):

  • Zinc(Zn) is a vital element that our systems need to absorb food and nutrients as well as create healthy skin and bones. Zinc(Zn) ions play a crucial role in a number of the body's enzymes.
  • Sulfur(S) is a pale yellow, tasteless, brittle solid that is also necessary for life. It is found in many proteins as well as the amino acids cysteine and methionine. It is a trace element found in bone minerals, bodily fluids, and lipids.

Chemical reaction -

In this experiment, heating a zinc(Zn) and sulfur(S) combination causes an interesting chemical reaction. A blinding flash of light, hot sparks, a hissing sound, and a cloud of white smoke in the shape of a mushroom are produced.

Therefore, the following chemical processes are taking place in the reaction: Zn (s) + S (s) = ZnS (s).

Learn more about Zinc(Zn) here:

brainly.com/question/13890062

#SPJ1

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A sound wave traveling through dry air has a frequency of 15 Hz, a
Korolek [52]

Answer:

Option B

Explanation:

Speed of a wave is denoted by:

v=fλ

where f is the frequency which is unchanged 15Hz and λ is the new wavelength which is 28m

v=fλ

v=15(28)\\v=420m/s

3 0
3 years ago
Read 2 more answers
if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given
yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given,  y=80t-16t^2 gives the height in 't' seconds.

Average velocity = Rate of change of distance

                             = Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1)  t = 0.01 s

    y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

    Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

    y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

    Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by  y=80t-16t^2 .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

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7 0
2 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
Which organ system provides structural support to other organ systems
shutvik [7]

Answer:

<h3>A. Skeletal </h3>

Explanation:

I hope it helps ❤❤

3 0
2 years ago
After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter
Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, \Delta L=3.1\ cm

(a) The change in length per unit original length is called strain. So,

\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193

(b) The force acting per unit area is called stress.

\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa

(c) The ratio of stress to the strain is called Young's modulus. So,

Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2

Hence, this is the required solution.

8 0
3 years ago
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