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zhuklara [117]
3 years ago
5

Why did Rutherford choose alpha particles in his experiment?

Chemistry
1 answer:
Georgia [21]3 years ago
5 0
Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.
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How many atoms are in 25.00 g of B?
klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number (N_A). This number is equivalent to 6.022 \times 10^{23}.

This number can be used to convert any values to:

  • atoms
  • molecules
  • formula units
  • moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

<u>1. Locating the number of moles of B in the sample</u>

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

  • Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

  • \displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}
  • \displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

  • \displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}

We can now simplify the expression. Our <u>grams B</u> unit will cancel out, so we are therefore left with <u>moles B</u> remaining.

<u>2. Locating the number of atoms in the sample</u>

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

  • \displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}
  • \displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

  • \displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}

This expression can now be operated. You will need a calculator to perform this calculation.

<u>Our numerator is:</u>

  • [(25.00 \times 1 \times (6.022 \times 10^{23})]

Plugging this into a calculator, we get:

  • 1.5055 \times 10^{25}

<u>Our denominator is:</u>

  • (1 \times 10.81 \times 1)

This simplifies to:

  • 10.81

<u>Dividing our numerator and denominator:</u>

  • <u />\displaystyle \frac{1.5055 \times 10^{25}}{10.81}<u />

Plugging this into a calculator, we get:

  • 1.392691952 \times 10^{24}

<u>3. Simplifying with significant figures</u>

Now, we need to take into account that we have significant figures. We are given this original value:

  • 25.00

This value has four significant figures, which means we need to round our value we received above to four significant figures.

  • \approx 1.393

Our units are added as well as our scientific notation:

  • 1.393 \times 10^{24} \ \text{atoms of B}

Therefore, our final answer is choice A.

8 0
3 years ago
The standard on which the atomic mass unit is based is the mass of a
pashok25 [27]

Answer;

D. carbon-12 atom

Explanation;

-One atomic mass unit is the mass that is exactly equal to one-twelfth the mass of one carbon-12 atom. Therefore, the mass of carbon-12 is 12 amu and this provides the standard that is used in measuring the atomic mass of all other elements.

-The atomic mass of carbon is 12.01 and not 12.00 in the periodic table.

-Due to the fact that there are more than one element for most naturally occurring elements, the average mass of the naturally occurring mixture of isotopes is taken while determining the atomic mass of an element.

3 0
3 years ago
Bohr's planetary of the atom stated that
adelina 88 [10]
He proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another.
3 0
3 years ago
If you visited a virtual Newton’s Cradle website or played with the real device, you might have noticed that after a few swings,
vesna_86 [32]

Answer:

gravite

Explanation:

8 0
3 years ago
How many valence electrons does copper contain?
iren [92.7K]
Copper has 29 electrons, and its electron configuration is:
1s2 2s2 2p6 3s2 3p6 3d10 4s1.

Therefore, copper has 1 valence electron.

Hope this helps~


8 0
4 years ago
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