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3 years ago

What type of reaction is shown in the following chemical equation: 2H2O → 2H2 + O2?

Physics

2 answers:

Kay [80]3 years ago

4 0

Answer:

Decomposition

Explanation:

A reaction in which one compound breaks up into two or more compounds is called a decomposition reaction.

atroni [7]3 years ago

4 0

Answer:

It is a decomposition reaction.

Explanation:

In a decomposition reaction, a single substance breaks down or breaks down, producing two or more different substances. In other words, decomposition reactions are those that form two or more substances from a compound. In this reaction the atoms that form a compound are separated to give the products according to the formula:

AB → A + B

This type of reactions can occur spontaneously or caused by certain external factors, such as heat.

In this case, you will see that the H₂O molecule decomposes to form two substances: H₂ and O₂. Then you will see that it is a decomposition reaction.

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A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0

2 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

3 years ago

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

kotykmax [81]

Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

5 0

3 years ago

Read 2 more answers

A B-52 bomber jet flies at a horizontal velocity of 286.2 m/s and at an altitude of 7500 m above the ground.

kykrilka [37]

Answer:

you need to divide 7500/286.2

3 0

2 years ago

A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0

3 years ago