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2 years ago

To stop an object that is in motion on a frictionless surface, you stop applying a force.

Chemistry

1 answer:

madreJ [45]2 years ago

8 0

The correct answer is true, you need to be able to stop it, an object in motion remains in motion unless acted upon

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Read the following passage.

Archy [21]

Answer is: the discovery of sub atomic particles like electrons.

J. J. Thomson discovered the electron in 1897.

His "plum pudding" model (1904) suggested: the electrons are embedded in the positive charge.

With this model, he abandoned his earlier hypothesis (the atom was composed of immaterial vortices).

J.J. Thomson placed two oppositely charged electric plates around the cathode ray. He did experiments using different metals as electrode materials and found that the properties of the cathode ray remained constant no matter what cathode material he used.

Tomson concluded that atoms are divisible and that the corpuscles are their building blocks (atoms are made up of smaller particles).

7 0

2 years ago

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Hypothesis: If a material undergoes a

zepelin [54]

Answer:

I hate to not answer and have you repost this if you could repost it with the choices by clicking the arrow I can figure it out a lot faster and I'll copy and paste to show you that it's right

Explanation:

I'm good with history biology sum math so if you want to do what I asked and reposted I can give you the answers and I will show that they are correct I won't just guess like some people do just to get points cuz I don't care about points I just get on here to help people

8 0

2 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$