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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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This kind of bond is formed when an atom with low electronegativity donates electrons to an
Butoxors [25]

Answer:

Ionic Bond

Explanation:

The atom with the higher electronegativity wants to fill its valence electron shell (meaning it wants 8 electrons in this shell). The atom with lower electronegativity will want to empty <em>or donate </em>an electron so that it can have an empty valence shell.

3 0
3 years ago
What is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?
zaharov [31]

Answer:

<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>

Explanation:

Since CH _ 3COOH is a weak acid to find the pH of CH _ 3COOH we use the formula

pH =  -  \frac{1}{2}   log(Ka)  -  \frac{1}{2}  log(c)

where

Ka is the acid dissociation constant

c is the concentration

From the question

Ka of CH _ 3COOH = 1.75 × 10^-5

c = 1.00 × 10-³M

Substitute the values into the above formula and solve for the pH

That's

pH =  \frac{1}{2} ( -  log(1.75 \times {10 }^{ - 5} -  log(1.00 \times  {10}^{ - 3} )  )  \\   =  \frac{1}{2} (4.757  + 3) \\  =  \frac{1}{2}  \times 7.757) \\  = 3.8785 \:  \:  \:  \:  \:  \:  \:  \:

We have the answer as

<h3>pH = 3.9</h3>

To find the pOH we use the formula

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.9

We have the answer as

<h3>pOH = 10.1</h3>

Hope this helps you

4 0
3 years ago
An archer pulls her bowstring back 0.370 m by exerting a force that increases uniformly from zero to 264 N. (a) What is the equi
Arte-miy333 [17]

Answer:

713.51 N/m

Explanation:

Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.

From hook's law,

F = ke ...........................Equation 1

Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.

Make k the subject of the equation,

k = F/e ............................ Equation 2

Given: F = 264 N, e = 0.37 m.

Substitute into equation 2

k = 264/0.37

k = 713.51 N/m

Hence the spring constant of the bow  = 713.51 N/m

3 0
2 years ago
If 14 moles of oxygen react with 14 moles of hydrogen to produce water, what is the
Leokris [45]
Hydrogen is the limiting reactant because when doing a stoichiometry equation for the reactants, hydrogen will be used completely by having a smaller yield and oxygen will be excess (7 moles to be exact)

7 0
2 years ago
calculate δg o for each reaction using δg o f values: (a) h2(g) i2(s) → 2hi(g) 2.6 kj (b) mno2(s) 2co(g) → mn(s) 2co2(g) kj (c)
Reika [66]

(a)The change in Gibbs free energy for the reaction has been 2.6 kJ/mol.

(b) The change in Gibbs free energy for the reaction has been -49.3 kJ/mol.

(c) The change in Gibbs free energy for the reaction has been 91.38 kJ/mol.

6 0
2 years ago
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