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Aleonysh [2.5K]
2 years ago
8

The amount of oxygen in the reactants is 4 atoms. In the products, the oxygen is distributed to water (H2O) and O2 gas. Which co

efficients would be placed in front of water and oxygen?
2 in front of water and 2 in front of oxygen
1 in front of water and 2 in front of oxygen
1 in front of water and 1 in front of oxygen
2 in front of water and 1 in front of oxygen
Physics
2 answers:
Reil [10]2 years ago
6 0

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

This question is describing balancing a chemical reaction. A balanced chemical reaction has the same number of atoms of each elements on both the reactant and product side. According to the question, the reactants contains 4 atoms of oxygen. The reactants give rise to water (H20) and O2 in the products side.

This reaction is most likely the decomposition of hydrogen peroxide (H2O2) as follows:

H2O2 (l) ----> H2O (l) + O2(g)

Based on the description, H2O2 will be 2H2O2 as it is said to contain four atoms of oxygen. This means that, in order to have a balanced equation, we must place coefficient 2 in front of water and coefficient 1 in front of oxygen. That is;

2H2O2 (l) ----> 2H2O (l) + O2(g)

Harman [31]2 years ago
5 0

Answer:

Answer:

2 in front of water and 1 in front of oxygen

Explanation:

This question is describing balancing a chemical reaction. A balanced chemical reaction has the same number of atoms of each elements on both the reactant and product side. According to the question, the reactants contains 4 atoms of oxygen. The reactants give rise to water (H20) and O2 in the products side.

This reaction is most likely the decomposition of hydrogen peroxide (H2O2) as follows:

H2O2 (l) ----> H2O (l) + O2(g)

Based on the description, H2O2 will be 2H2O2 as it is said to contain four atoms of oxygen. This means that, in order to have a balanced equation, we must place coefficient 2 in front of water and coefficient 1 in front of oxygen. That is;

2H2O2 (l) ----> 2H2O (l) + O2(g)

Explanation:

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Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?
Maurinko [17]

Distance traveled by the ball is given by

distance = speed \times time

here we know that

speed = 20 m/s

times = 0.25 s

now we have

distance = 20 \times 0.25

distance = 5 m

so ball will travel 5 m distance in the given interval of time

6 0
3 years ago
What are two ways to increase impulse?
Pepsi [2]

Answer:

increase the time the force acts or you could increase the number of temptations.

hope this helped!

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sasho [114]

I believe the acceleration would be 5m/s

All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.

3 0
3 years ago
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guapka [62]
Dinosaurs but I need the whole groups yo tell you ;)
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