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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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Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

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What do group 2 elements have in common
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A seagull flies at a velocity of 9.00 m/s straight into the wind.
RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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2 years ago
According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?
KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

4 0
3 years ago
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Urgent need help 100 points
Sati [7]

Answer: The velocity with which the sand throw is 24.2 m/s.

Explanation:

Explanation:

acceleration due to gravity, a =  3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial  velocity at the time of throw is u.

Use third equation of motion

The velocity with which the sand throw is 24.2 m/s.

7 0
2 years ago
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