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forsale [732]
3 years ago
10

Which set of elements have the most similar chemical properties?

Chemistry
1 answer:
EleoNora [17]3 years ago
7 0

answer: A not sure but i hope it helps :) success

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How many liters of air must react with 2500 mL of hexane in order for combustion to occur completely. The percentage of oxygen i
bagirrra123 [75]

Answer:

118750 ml

Explanation:

The chemical equation for complete combustion of hexane is given as;

2C6H14 + 19O2 → 12CO2 + 14H2O  

From the equation of the reaction;

2 mol of C6H14 reacts with 19 mol of O2

2 ml of C6H14 reacts with 19 ml of O2

2500 mL of C6H14 would react with x ml of O2

2 = 19

2500 = x

x = 2500 * 19 / 2 = 23750 ml

Since oxygen is 20% of air;

23750 = 20 / 100 * (Volume of air)

Volume of air = 23750 * 100 / 20 = 118750 ml

7 0
3 years ago
Suppose a student did not fully redissolve the Al(OH)3 that initially formed upon the addition of H2SO4. Predict and explain how
kirza4 [7]
<span>Percent yield is the measured recovery of a substance in grams divided by the theoretical amount of substance expected in grams (times 100 to convert to %). Undissolved Aluminum trihydroxide would add mass to the Aluminum sulphate one is making here. The measured mass would be inaccurately high, increasing the percent yield calculated.</span>
6 0
3 years ago
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
Could you please show work if you could
vovangra [49]

Answer:

can u writ qn

Explanation:

6 0
3 years ago
Draw an atomic model from1 to 20 elements with their symbol valency and electronic configuration​
ale4655 [162]

Answer:

MARK AS THE BRAINLIEST ANSWER

Explanation:

BRAINLIEST

3 0
3 years ago
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