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Oksanka [162]
3 years ago
14

Approximately how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

Chemistry
2 answers:
Ann [662]3 years ago
8 0
<span>c. 30 times The Richter scale is a logarithmic scale for the energy release during an earth quake. For every increase by 1 in magnitude, the energy released is about 31.6 times larger. So calculate the magnitude difference by subtracting. 6.5 - 5.5 = 1.0 And then multiply by 31.6, giving 31.6. And since we're just looking for approximate values, select the answer that's closest which is option "c. 30 times"</span>
Elanso [62]3 years ago
4 0
Answer: (C) 30 times

Richter scale is used to determine the magnitude of an earthquake from the arrival time of P and S waves. It determines the total amount of energy released during an earthquake.
If Richter magnitude scale measures 6.5 then it produces 30 times more energy than 1 Richter scale magnitude during an earthquake. The magnitude of an earthquake is plotted on a logarithmic scale from 1 to 9 and increases 10 folds by one magnitude. If 6.5 is the magnitude, it means it is 10 times more than 5.5 magnitude earthquake. Similarly the amount of energy is calculated as 30 folds. So 6.5 magnitude will have 30 folds more energy released than that with 5.5 magnitude earthquake.
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Calculate the percent by mass of 10 grams of calcium hydroxide mixed in 65 grams of water
sukhopar [10]

Answer:

13.3 %

Explanation:

Percent by mass indicates the mass of solute in 100g of solution.

We must know the solution's mass:

Solution's mass = Solute mass + solvent mass

Solution's mass = 10 g + 65 g = 75 g

Percent by mass = (10 g / 75 g).100 = 13.3 %

Otherwise, we can make a rule of three:

In 75 g of solution we have 10 g of solute

In 100 g of solution we must have (100 . 10) /75 = 13.3 %

5 0
3 years ago
List at least three things that your body strives to keep constant. (9 points) Think about what
Evgen [1.6K]

Answer:

Temperature, heart rate and blood flow.

Explanation:

5 0
3 years ago
From the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆ , find the number of steps needed to produce a sample of the enriched f
Dafna11 [192]

The number of steps required to manufacture a sample of the 3.0 mole%  ²³⁵U enriched fuel used in many nuclear reactors from the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆. ²³⁵U occurs naturally in an abundance of 0.72% are :  mining, milling, conversion, enrichment, fuel fabrication and electricity generation.

<h3>What is Uranium abundance ? </h3>
  • The majority of the 500 commercial nuclear power reactors that are currently in operation or being built across the world need their fuel to be enriched in the U-235 isotope.
  • This enrichment is done commercially using centrifuges filled with gaseous uranium.
  • A laser-excitation-based method is being developed in Australia.
  • Uranium oxide needs to be changed into a fluoride before enrichment so that it can be treated as a gas at low temperature.
  • Uranium enrichment is a delicate technology from the perspective of non-proliferation and needs to be subject to strict international regulation. The capacity for world enrichment is vastly overbuilt.

The two isotopes of uranium that are most commonly found in nature are U-235 and U-238. The 'fission' or breaking of the U-235 atoms, which releases energy in the form of heat, is how nuclear reactors generate energy. The primary fissile isotope of uranium is U-235.

The U-235 isotope makes up 0.7% of naturally occurring uranium. The U-238 isotope, which has a small direct contribution to the fission process, makes up the majority of the remaining 99.3%. (though it does so indirectly by the formation of fissile isotopes of plutonium). A physical procedure called isotope separation is used to concentrate (or "enrich") one isotope in comparison to others. The majority of reactors are light water reactors (of the PWR and BWR kinds) and need their fuel to have uranium enriched by 0.7% to 3-5% U-235.

There is some interest in increasing the level of enrichment to around 7%, and even over 20% for particular special power reactor fuels, as high-assay LEU (HALEU).

Although uranium-235 and uranium-238 are chemically identical, they have different physical characteristics, most notably mass. The U-235 atom has an atomic mass of 235 units due to its 92 protons and 143 neutrons in its nucleus. The U-238 nucleus has 146 neutrons—three more than the U-235 nucleus—in addition to its 92 protons, giving it a mass of 238 units.

The isotopes may be separated due to the mass difference between U-235 and U-238, which also makes it possible to "enrich" or raise the proportion of U-235. This slight mass difference is used, directly or indirectly, in all current and historical enrichment procedures.

Some reactors employ naturally occurring uranium as its fuel, such as the British Magnox and Canadian Candu reactors. (By contrast, to manufacture at least 90% U-235, uranium needed for nuclear bombs would need to be enriched in facilities created just for that purpose.)

Uranium oxide from the mine is first transformed into uranium hexafluoride in a separate conversion plant because enrichment operations need the metal to be in a gaseous state at a low temperature.

To know more about Effusion please click here : brainly.com/question/22359712

#SPJ4

7 0
2 years ago
A heterogeneous mixture that has very small dispersed particles and stays mixed for a long time is a____. it for science class
Shkiper50 [21]
The heterogeneous mixture that has very small dispersed particles and stays mixed for a long time is colloid. Because colloid has particles that are small enough to suspended but are as large that they can scatter light.
8 0
3 years ago
Rachel burns a 53 gram cracker under a soda can filled with 81.6 grams of water. She took the temperature of the water before sh
Anna11 [10]

Answer

Q=4479.8 cal

Procedure

To solve the problem you will need to use the specific heat formula

Q=mc\Delta T

Where;

Q=heat energy

m=mass

c=specific heat capacity

ΔT=change in temperature

Assuming that the heat released from the cracker of unknown material is equal to the heat absorbed by the water, then we can use the c and m for water in our calculations.

c_{water}=4.186\frac{J}{g\degree C}

Substituting the values in our equation we have

Q=mc\Delta T=81.6\text{ }g(4.186\frac{J}{g\degree C})(68.1-13.2)\degree C=18752.61\text{ }J

Finally, transform the J to cal

18752.61\text{ }J\frac{1\text{ }cal}{4.186\text{ }J}=4479.8\text{ cal}

8 0
1 year ago
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