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ikadub [295]
3 years ago
12

Why do you think that scientists in the 20th century thought that it was important to “unlock” the structures of the atom?

Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

The 20th century saw huge advances in our understanding and use of the nucleus. For instance, in 1939 scientists Otto Hahn, Lise Meitner and Otto Frisch discovered nuclear fission – a process by which radioactive materials release energy when they are induced to split.

Realising the huge amount of energy that such a reaction produces, scientists were tasked with developing this new knowledge initially for harm in nuclear weapons. Just six years after fission’s discovery, it was harnessed in the atom bombs that destroyed the Japanese cities Hiroshima and Nagasaki, and controversially ended the Second World War. Later, much more powerful hydrogen bombs were developed that combined fission with the process powering the Sun – fusion.

Hope this helps! PLEASE GIVE ME BRAINLIEST!!!!! =)

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How many molecules are there in 79g of Fe2O3? how many atoms is this?
Leya [2.2K]
There are approximately 160 grams in 1 mol of Fe2O3 molecules. Therefore, there would be 79/160= 0.49375 mols of Fe2O3 molecules in 79 grams. There are 5 atoms in total for each molecule of Fe2O3, therefore 79/160 * 5 = 79/32 = 2.46875 mols of atoms.
4 0
3 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
A student weighs out 3.68 grams of cobalt (II) chloride to do an experiment with. How many moles of is this?
leva [86]

Answer:

0.029mol

Explanation:

from

n=m÷mr

where

n is number of moles

m is mass

mr is molar mass

8 0
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Which of these is a TRUE statement? A) Sound travels through gases easier than liquids. B) Sound travels through solids easier t
bekas [8.4K]

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3 years ago
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What is the final temperature when 60.2 g of water at 72.4°C is mixed with 20.3 g of water at 34.4°C? (The specific heat of wate
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Answer:

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Explanation:the celcuis would over power the fairnheight

5 0
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