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Paraphin [41]
2 years ago
8

How many grams of oxygen are needed to react with excess methane in the following reaction, in order to produce 325 kJ of energy

?
Chemistry
1 answer:
Sergio [31]2 years ago
6 0
~移動NDISとJSが可いままでも今のところ天の
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Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was react
stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
8 0
3 years ago
Which substance has a definite shape and definite volume?
tino4ka555 [31]

Answer:

D iced tea

Explanation:

right answers

3 0
2 years ago
The purpose of the salt bridge in an electrochemical cell is to ____.
IgorLugansk [536]

Answer:

A. maintain electrical neutrality in the half-cells via migration of ions

Explanation:

Salt bridge -

For an electrochemical reaction , involving an anode and a cathode , both the electrodes are connect via a salt bridge to complete the circuit for the reaction .

One of the very important use of a salt bridge is to maintain the electrical neutrality of the respective half cells , which is achieved by the movement of ions .

Hence , from the given options , the correct option is ( a ) .

8 0
3 years ago
Why does the sun move from night-time to dayime?
valentina_108 [34]

Answer: Rotation cause by the earths orbit and gravity

Explanation:

4 0
2 years ago
Read 2 more answers
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0
3 years ago
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