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VikaD [51]
3 years ago
9

What are the two most common ways to produce hydrogen gas used in fuel cells?

Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

Steam-methane reforming

Electrolysis of water

Explanation:

Steam methane reforming involves reaction of methane with water in the presence of a catalyst such as nickel to form Carbon oxides and Hydrogen.

CH4 + H2O ⇌ CO + 3 H

Electrolysis of water involves splitting of water through application of electric current to give Hydrogen and Oxygen gas.

2 H2O(l) → 2 H2(g) + O2(g)

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What do we call the certain electrons that are able to sometimes wander<br> away or be shared? *
Natalka [10]

Answer:

compond

Explanation:

5 0
3 years ago
Sasha did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers
chubhunter [2.5K]

Answer:

b is the higher solubility then A

4 0
3 years ago
While skateboarding at 19 km/h throwning a tennis ball at 11 km/h what is the speed of the ball
Lina20 [59]

According to whom ?

So YOU're on your skateboard, and there's somebody else, sitting on HIS porch, watching you skate by on your board.

-- The man on the porch says you're skating by him at 19 km/hr .

-- You throw a tennis ball.  

. . . . . Do you throw it in the same direction that you're skateboarding, or do you throw it away behind you, toward the place you just came from ?

. . . . . Does it fly away from YOU at 11 km/hr ?  Or does it fly past the man on the porch at 11 km/hr ?

There are 4 possible combinations.  One of them is not possible.  Each of the other three combinations leads to two different answers to the question.  And ALL six answers are correct !

1).  You throw the ball forward, in the same direction you're skating.  It flies away from your hand at 11 km/hr.

To you, the speed of the ball is 11 km/hr, in the direction you're skating.  To the man on the porch, it's 30 km/hr, in the direction you're skating.

2). You throw the ball forward, in the same direction you're skating.  It flies past the porch at 11 km/hr.

This isn't possible.

3). You throw the ball backward, toward where you just came from.  It flies away from YOU at 11 km/hr.

To you, the speed of the ball is 11 km/hr, in the direction backward from you.  To the man on the porch, the speed of the ball is 10 km/hr in the direction you're skating.

4).  You throw the ball backward, toward where you just came from.  It flies past the porch at 11 km/hr.

To you, the speed of the ball is 8 km/hr, in the direction backward from you.  To the man on the porch, it's 11 km/hr in the direction you're skating.


NOW you're going to ask me "But what's the REAL speed of the ball ?"

The answer to THAT one is:  There's no such thing !  It all depends on WHO's measuring it ... where that observer is and how HE's moving.

The displacement, speed, velocity, acceleration, and energy of the ball, ALL depend on who's watching it and measuring it.

I'll be interested to see whether you mark this answer 'Brainliest', or report it because it's weird, confusing, and ridiculous.

3 0
4 years ago
When you and a friend move a couch to another room, you exert a force of 75 N over 5 m. How much work did you do?
bagirrra123 [75]

Answer:375J

Explanation:

Force=75N

Distance=5m

Work=force x distance

Work=75 x 5

Work=375J

5 0
3 years ago
Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler
ehidna [41]

A) Acceleration in part A: 6.1 m/s^2

B) Acceleration in part B: 2.7 m/s^2

C) Acceleration in part C: 1.5 m/s^2

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

\Delta t = 0.9 - 0 = 0.9 s

So the acceleration is

a=\frac{5.5-0}{0.9}=6.1 m/s^2

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

\Delta t = 2.4 - 0.9 = 1.5 s is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

a=\frac{9.5-5.5}{1.5}=2.7 m/s^2

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

\Delta t = 3.4 - 2.4 = 1 s is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

a=\frac{11-9.5}{1}=1.5 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
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