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Vlad [161]
3 years ago
10

Which statement best expresses Newton's first law of motion? Group of answer choices Objects are constantly speeding up or slowi

ng down. All objects resist changes to their states of motion. For every action, there is an equal and opposite reaction. The forces that cause objects to move will decrease over time.
Chemistry
1 answer:
nadezda [96]3 years ago
7 0

Answer:

All objects resist changes to their states of motion.  

Explanation:

Sir Issac Newton was an English physicist, mathematician, astronomer and a well known author. Besides, he was a great scientist. Newton discovered many scientific phenomenon and scientific theories in nature.

The most important and famous discoveries of Newton is the Newton's laws of motion. Newtons stated three laws of motion, namely, Newtons 1st law of motion, Newton's 2nd laws of motion and Newton's 3rd law of motion.

According to Newtons's 1st law of motion : A body continues to be in the state of motion or in the state of rest until and unless an external force is applied to it. In other words, all bodies resists changes to the states of their motion or rest.

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3 years ago
Describe light with respect to its speed and its dual nature as both a wave and a particle.
Sedbober [7]

Answer:

Scientists have been debating over light being a wave or particle since its recognition.

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3 years ago
5. Identify each of the following as an element, a mixture, or a compound.
VashaNatasha [74]

Answer:

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4 0
3 years ago
Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

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