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snow_tiger [21]
2 years ago
10

En 1500 kg de aire hay 0.8 g de dióxido de carbono a cuantos ppm corresponde la muestra de co2

Chemistry
1 answer:
Alex Ar [27]2 years ago
4 0

Translation:

In 1500 kg of air there are 0.8 carbon dioxide. How many ppm does the carbon dioxide sample correspond to?

Explanation:

Given parameters:

Mass of air = 1500kg

Mass of carbon dioxide= 0.8g

Unknown:

PPM of carbon dioxide in the sample = ?

SOLUTION :

The parts per million ppm is used to express dilute concentration of solutes. It simple means out of a million.

To express the answer in ppm, first convert into uniform units:

  1500kg to g = ?

             1kg = 1000g

             1500kg = (1500 x 1000)g = 1500000g

To express as part per million:

\frac{mass of carbon dioxide}{mass of air} x 1000000

         \frac{0.8}{1500000} x 1000000 = 0.533ppm

Learn more:

Parts per million brainly.com/question/2854033

#learnwithBrainly

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Answer:

Ok so,  b. A redox reaction occurs in an electrochemical cell, where silver (Ag) is oxidized and nickel (Ni) is reduced - In voltaic cells, also called galvanic cells, oxidation occurs at the anode and reduction occurs at the cathode. A mnemonic for this is "An Ox. Red Cat." So since silver is oxidized, the silver half-cell is the anode. And the nickel half-cell is the cathode...

i. Write the half-reactions for this reaction, indicating the oxidation half-reaction and the reduction half-reaction- The substance having highest positive  potential will always get reduced and will undergo reduction reaction. Here, zinc will always undergo reduction reaction will get reduced

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iii. Calculate the standard potential (voltage) of the cell

Look up the reduction potential,

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Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction,

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iv. What kind of electrochemical cell is this? Explain your answer.

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3 0
2 years ago
What is the maximum amount of CO2
Wittaler [7]

Answer:

10.6 g CO₂

Explanation:

You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.

To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*

C₂H₂ + O₂ ----> 2 CO₂ + H₂

9.31 g C₂H₂            1 mole               2 moles CO₂          44.0095 g
------------------  x  -------------------  x  ----------------------  x  -------------------  =
                            26.0373 g           1 mole C₂H₂              1 mole

=  31.5 g CO₂

3.8 g O₂             1 mole               2 moles CO₂          44.0095 g
-------------  x  --------------------  x  ----------------------  x  --------------------  =
                       31.9988 g              1 mole O₂                 1 mole

=  10.6 g CO₂

10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.

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Answer:

The new pressure will be 0.225 kPa.

Explanation:

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P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

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Putting values in above equation, we get:

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