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baherus [9]
4 years ago
13

In Rutherford’s gold foil experiment, most alpha particles passed through the gold foil without deflection and were detected on

the screen. What caused the particles to pass through without any deflection?
Chemistry
2 answers:
earnstyle [38]4 years ago
8 0

In Rutherford's gold foil experiment, some of those particles were deflected back towards him and off to the side. This told him that the protons and electrons were not evenly dispersed. However, the particles that did pass through did so because the protons are located in a concentrated area in the center of the atom, and the electrons are dispersed in the empty space around it. At least, that is what Rutherford determined

Komok [63]4 years ago
4 0

<u>Answer:</u> The empty space in the nucleus let the particles pass through without any deflection.

<u>Explanation:</u>

Rutherford gave an experiment which is known as gold foil experiment.

In this experiment, he took a gold foil and bombarded it with the alpha particles that carries positive charge of 2 units. He expected that the particles will pass straight through the foil, but to his surprise, some of them deflected their path and a few of them bounced back.

From this he gave a result that an atom consist of empty space and also has a small positive charge which is present at the center. Due to this positive charge, the alpha particles deflected their path and some of them bounced straight back from their path.

As, the atom consist of alot of empty space. Thus, the alpha particles passed straight through without any deflection.

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How many stable Lewis resonance structures does NO2- HAVE<br> 3<br> 1<br> 2<br> 4
Natalka [10]

Answer:

2

Explanation:

it has a total of 17 valence electrons

6 0
3 years ago
What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were
irinina [24]

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

NaF+HCl\rightarrow NaCl+HF

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

[HF]=\frac{0.450mol}{1.0L} =0.450M

[NaF]=\frac{0.050mol}{1.0L} =0.050M

Now, calculated the pKa of HF:

pKa=-log(6.8x10^{-4})=3.17

We can proceed to the HH equation:

pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21

Best regards!

6 0
3 years ago
Why does a balanced chemical equation support the law of conservation of mass
Elena L [17]

Answer:

A matter can not be created or destroyed in chemical reactions. In every chemical reactions, the same mass of matter must end up in the product as started in the reactants

8 0
3 years ago
There are exactly 2.54 centimeters in 1 inch. When using this conversion factor, how many 20) significant figures are you limite
HACTEHA [7]

Answer : The correct option is, (E) infinite number of significant figures.

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.

All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.

All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.

All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.

All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

The given conversion is:

2.54 cm = 1 inch

2.54 has 3 significant figure.

1 has infinite number of significant figures.

Hence, the correct option is, (E) infinite number of significant figures.

5 0
3 years ago
A. Neutrons and electrons
Marianna [84]

Answer:

D.

Explanation:

Because protons and neutrons make up the mass of an atom.

5 0
3 years ago
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