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djverab [1.8K]
2 years ago
11

Gas Law- please help!!!

Chemistry
1 answer:
jekas [21]2 years ago
8 0

The pressure of the carbon dioxide will be 0.09079 atm.

<h3>What is partial pressure?</h3>

The pressure exerted by the individual gas is known as partial pressure.

The partial pressure is given as

\rm P_{Total} = P_1 + P_2 + ...

In a mixture of carbon dioxide and oxygen, 40.0% of the gas pressure is exerted by oxygen.

If the total pressure is 115 mmHg.

The total pressure in atm will be

P = 115 mmHg

P = 0.15132 atm

We have

\rm P_T \ \  \  = P_{O_2} + P_{CO_2}\\\\0.15132 \ \  = 0.4 \times 0.15132 +P_{CO_2}\\\\ P_{CO_2} = 0.09079\ atm

Then the pressure of the carbon dioxide will be 0.09079 atm.

More about the partial pressure link is given below.

brainly.com/question/13199169

#SPJ1

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When CH₄ is burnt in excess O₂ following products are formed,

                           CH₄  +  2 O₂     →     CO₂  +  2 H₂O

According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,

                          1 mole of CO₂  and  2 moles of H₂O

Converting 1 mole CO₂ to grams;
As,
                           Mass  =  Moles × M.mass

                           Mass  =  1 mol ×  44 g.mol⁻¹

                           Mass  =  40 g of CO₂

Converting 2 moles of H₂O to grams,

                           Mass  =  2 mol ×  18 g.mol⁻¹
                         
                           Mass  =  36 g of H₂O

Total grams of products;

                           Mass of CO₂  =  44 g
                    +     Mass of H₂O  =  36 g
                                                  -------------
                           Total              =   80 g of Product

Result:
            80 grams of product
is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
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If a combination reaction takes place between rubidium and bromine, the chemical formula for the product is
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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
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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.

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Liquid EO                       -77.4

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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)

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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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