CHEMISTRY HIGH SCHOOL

During nuclear fission, great amounts of energy are produced from____. A. particle accelerators B. tremendous amounts of mass C. a series of chemical reactions D. very small amounts of mass.

Answers

Answer 1
Answer: During nuclear fission, great amounts of energy are produced from____.

Answer: Out of all the options presented above, the one that completes the statement and makes it true, is answer choice D) very small amounts of mass. Nuclear fission power does not produce carbon and also produces 12% of the world's power.

I hope it helps, Regards.

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A fossilized animal that was killed by a volcanic eruption has 6.25 percent of its original amount of carbon-14 remaining; this radioisotope has a half-life of 5730 years. How long ago did the volcano erupt?a) 34,380 years agob) 22,920 years agoc) 17,190 years agod) 9,168 years ago

Answers

With the assumption that the animal is 100kg the remaining mass is 6.25. (assumption of mass will not matter since the decrease in mass is in percentage.) 

To answer this we use the formula A=Aoe^-(ln2/t1/2)t
where 
A- final amount (6.25kg)
Ao - initial amount (100kg)
t1/2 - half life (5730)
t - time it takes to change A to Ao

plugging in the values 
t= 22920 years ago. 

What is the name of the movement of the Earth around the Sun as pointed out by letter J.

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it is called the Earth's rotation

Sand + water = Sand Sand solution​

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Answer:

When sand is added to water, it does not mix, and no solution is formed.

Explanation:

Sand is insoluble...

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3(s)? 2NH3(g) + 2O2(g) NH4NO3(s) + H2O(l)Given:
NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K.
NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K.
H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K.
O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K.

Answers

∆G =  ∆H - T∆S

NH3: ∆G = -46.11x10^3 - (298)(192.45) = -103460.1 J
O2: ∆G = 0 - (298)(205) = -61090 J
NH4NO3: ∆G = -365.56x10^3 - (298)(151.08) = -410581.84 J
H2O: ∆G = -285.830x10^3 - (298)69.91) = -306663.18 J

∆Grex = ∆Gproducts - ∆Greactants
∆Grex = (-410581.84 +  -306663.18) - (-103460.1/2 + -61090/2)
∆Grex =-634969.97 J/mol = -634.97 kJ/mol

Balance the equation :

...NO2 + ...O2 + ...H2O = ...HNO3

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We are going to do this by means of electronic balance:
N(IV)\Rightarrow\ N(V)+e^-|*4\n 2O(0)+4e^-\Rightarrow\ 2O^2^-\n 4NO_2+O_2+2H_2O\Rightarrow4HNO_3

If you hadn't learnt it yet, you would do it normally:
Count oxygens. To have even number of oxygen write "2" in fronf of H2O. Then to have equal hydrogen you have to add "4" in front of HNO3 and then NO2.

Which of the following devices can measure only atmospheric pressure?

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The answer is a barometer
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