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Anika [276]
2 years ago
13

How many molecules of Oxygen gas are there on the reactant side of this equation?

Physics
1 answer:
Sergeeva-Olga [200]2 years ago
8 0

Answer:

4

Explanation:

It has 8 O atoms and 4 O2(g) molecules

You might be interested in
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
Place least complex to most complex:
Reika [66]
Cell, tissue, organ, organ sytem, organism, population, community, ecosystem, biome. 
7 0
2 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?
EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by d = \frac { m } { v }.

Here in this problem, we are given a mass of 565 g which occupies a volume of 50 cm^3.

So plugging the data in the above formula to find the density:

Density = \frac { 565 } { 50 } = 11.3

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0
3 years ago
The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared
velikii [3]

Answer:

\displaystyle a(5)=-32

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

\displaystyle v(t)=\frac{df}{dt}

And the acceleration is

\displaystyle a(t)=\frac{dv}{dt}

Or equivalently

\displaystyle a(t)=\frac{d^2f}{d^2t}

The given height of a projectile is

f(t)=-16t^2 +238t+3

Let's compute the speed

\displaystyle v(t)=-32t+238

And the acceleration

\displaystyle a(t)=-32

It's a constant value regardless of the time t, thus

\boxed{\displaystyle a(5)=-32}

3 0
3 years ago
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