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ira [324]
3 years ago
9

What is anything that has mass and volume

Chemistry
2 answers:
Korvikt [17]3 years ago
6 0
Matter is what makes up all of the objects and living organisms in the universe. Anything that has mass takes up space. 

Pavlova-9 [17]3 years ago
3 0
Anything that has mass and volume (takes up space) is called matter.
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Which of these has the most covalent character? <br> NaF <br> Caf2 <br> Bef2 <br> CsF <br> SrF2
VLD [36.1K]
Bef2 is the compound that has the most covalent character. Whichever element has the greatest electronegativity has the least difference and most covalent character. 
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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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Answer:

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Explanation:

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I agree with the person on top have a beautiful day:)
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Tatiana [17]
To answer this problem, we must make assumptions for simplicity. The first assumption is that, the system only consist of these 3 gases. The second assumption is that, these gases behave ideally. Thus, from Dalton's Law of Partial Pressure, the total pressure is simply the sum of their individual partial pressures.

Total pressure = 2.5 + 0.8 + 3.4 = <em>6.7 atm</em>
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