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Naddika [18.5K]
3 years ago
9

Why did Rutherford say that bombarding atoms with particles was like “playing with marbles”?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
8 0

In 1917, Rutherford discovered something new, he bombarded alpha particles on nitrogen gas and noticed that occasionally an oxygen atom is produced. From this he concluded that the alpha particles removes proton from the nucleus (positively charged particle). He named this playing with marbles and he become the first one to split an atom. Thus, by bombarding nitrogen with alpha particles, he observed the production of oxygen, a different element.

Therefore, he said bombarding atoms with particles is similar to playing with marbles.

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Indicate type of chemical reactions for 2Mgl2+MN(SO3)2=2MgSO3+Mnl4<br>​
Sever21 [200]

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double decomposition reaction

7 0
3 years ago
Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production
Lynna [10]

Answer:

ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>

<em></em>

Thus <em>ΔH° of the reaction is -747.54kJ</em>

3 0
3 years ago
Determine the density of gold if a 475.09g sample occupy a space of 24.6cm^3
Ira Lisetskai [31]

Answer:

The answer is

<h2>19.31 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}

From the question

mass of gold = 475.09g

volume = 24.6 cm³

The density of the gold is

density =  \frac{475.09}{24 .6}  \\  = 19.3126016...

We have the final answer as

<h3>19.31 g/cm³</h3>

Hope this helps you

6 0
3 years ago
What is the total number of valence electrons in a carbon atom in the ground state
kolbaska11 [484]
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I hope this helps.  Let me know if anything is unclear.
6 0
2 years ago
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