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Advocard [28]
3 years ago
7

What is the balance equation for the chemical equation h2o+so2=h2o+s​

Physics
1 answer:
yaroslaw [1]3 years ago
7 0

Answer:

2H₂S + SO₂ —> 2H₂O + 3S

Explanation:

From the question given above, the following equation were obtained:

H₂S + SO₂ —> H₂O + S

The equation can be balance as illustrated below:

H₂S + SO₂ —> H₂O + S

There are 2 atoms of O on the left side and 1 atom on the right side it can be balance by writing 2 before H₂O as shown below:

H₂S + SO₂ —> 2H₂O + S

There are 2 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 2 before H₂S as shown below:

2H₂S + SO₂ —> 2H₂O + S

There are 3 atoms of S on the left side and 1 atom on the right side. It can be balance by writing 3 before S as shown below:

2H₂S + SO₂ —> 2H₂O + 3S

Now, the equation is balanced.!

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The center of gravity of a basketball is located _______.
zhenek [66]

Answer:

at its centre

Explanation:

The centre of gravity is a point inside the body where entire weight os the body is said to be concentrated. For geometrical symmetric bodies it lies at the geometric centre of the body.

The base ball is a spherical body, so its centre of gravity lies at its centre.

5 0
3 years ago
Select the correct answer.
soldi70 [24.7K]

Answer:

3

Explanation:

first you find the original component of the force be a cause in 36 degrees then with ka sin 36 find the average interforce by multiplying it with hundred Newton then divide the original force by the mass 25 Kg is equals tto ma

8 0
3 years ago
PLEASE HELP!!!!
Goryan [66]

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let u denote the initial velocity of the vehicle (20\; \text{mph} or 60\; \text{mph}) and let v denote the velocity of the vehicle after braking (0\; \text{mph}). Let x denote the braking distance.

Assume that the acceleration during braking are both constantly a in both scenarios. The SUVAT equations would apply. In particular:

\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}.

Since v = 0<em> </em>(the vehicle has completely stopped), the equation becomes x = (-u^{2}) / (2\, a).

Assuming that a (braking acceleration) stays the same, the braking distance x would be proportional to u^{2}, the square of the initial velocity.

Hence, increasing the initial speed from 20\; \text{mph} to 60\; \text{mph} would increase the braking distance by a factor of 3^{2} = 9.

7 0
1 year ago
Read 2 more answers
A car traveling at 38 m/s starts to decelerate steadily. It comes to a complete stop in 10 seconds. What is its acceleration?
asambeis [7]

a=v/t

a=38/10

a=-3.8 m/s^2

You put the negative in front of 3.8 because it decelerated.

Hope it helps and is correct :)

6 0
3 years ago
Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su
kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

F_g = \frac{GMm}{r^2}

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

7 0
3 years ago
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