How many inches are in 6.32 cm

Could Rutherford make any conclusions about electrons based on the result of experiment?

MatroZZZ [7]

Answer:

They are so small that they barely make up any of the mass of the atom, and they are so miniscule that during Rutherford 's Gold foil experiment ,they didn't even react with the alpha particles. They circle the nucleus on a ring. shown through Rutherford Atomic Model

hope it helps you

8 0

3 years ago

You are given three cubes, A, B, and C; one is magnesium, one is aluminum, and the third is silver. All three cubes have the sam

katovenus [111]

Answer:

The cube A is magnesium, the cube B is aluminum and the cube C is silver.

Explanation:

Density is defined by the expression $d=\frac{m}{V}$ where m is the mass and V is the volume, therefore:

- Density of the cube A:

$d_{A}=\frac{m_{A}}{V_{A}}$

- Density of the cube B:

$d_{B}=\frac{m_{B}}{V_{B}}$

- Density of the cube C:

$d_{C}=\frac{m_{C}}{V_{C}}$

Solving for mass:

$m_{A}=d_{A}*V_{A}$

$m_{B}=d_{B}*V_{B}$

$m_{C}=d_{C}*V_{C}$

And all the three cubes have the same mass, so:

$m_{A}=m_{B}=m_{C}$

Therefore:

$d_{A}*V_{A}=d_{B}*V_{B}$ (Eq.1)

$d_{A}*V_{A}=d_{C}*V_{C}$ (Eq.2)

Solving for $d_{1}$ in Eq.1:

$d_{A}=d_{B}\frac{V_{B}}{V_{A}}$

Replacing values for the volume:

$d_{A}=d_{B}\frac{16.7mL}{25.9mL}$

$d_{A}=d_{B}*0.64$

As we know the density of the aluminum is $2.7\frac{g}{cm^{3}}$ , so replacing this value for $d_{B}$ :

$d_{A}=2.7\frac{g}{mL}*0.64$

$d_{A}=1.728\frac{g}{mL}$

that is the density of the magnesium.

Solving for $d_{C}$ in Eq.2:

$d_{C}=d_{A}\frac{V_{A}}{V_{C}}$

$d_{C}=d_{A}\frac{25.9mL}{4.29mL}$

$d_{C}=d_{A}*6.04$

$d_{C}=1.728\frac{g}{mL}*6.04$

$d_{C}=10.4\frac{g}{mL}$

That is the density of the silver.

Therefore the cube A is magnesium, the cube B is aluminum and the cube C is silver.

8 0

4 years ago

If you purchased 0.610 μCi of sulfur-35, how many disintegrations per second does the sample undergo when it is brand new?Expres

ASHA 777 [7]

Answer:

the sample undergo $1.67 \times 10^5$ disintegrations per second.

Explanation:

Given:

The amount of sulphur purchased is 0.671 μCi

To find:

disintegrations per second = ?

Solution:

Some of the conversions are

$1 Ci = 3.8 \times 10^{10} Bq$

1 rad = 0.01 Gy

1Gy = 1 j/kg tissue

1 rem = 0.01 Sv

1Sv = 1 j/Kg

Using these conversions,

The decay rate of this sample is calculated as

$0.671 \mu C i \times \frac{1 \mu C i}{10^{6}} \mu C i \times \frac{0.671 \times 10^{10}}{1 \mu C i}$

= $1.67 \times 10^5$ disintegrations per second

7 0

4 years ago

Write and balance sodium when it burns in air to form sodium peroxide

Zanzabum

Answer:7.229 grams of oxygen is formed by the complete reaction of 35.23 g of metallic sodium with oxygen at 130–200 °C, a process that generates sodium oxide, which in a separate stage absorbs oxygen: 4 Na + O2 → 2 Na2O. The ozone oxidizes the sodium to form sodium peroxide.

7 0

3 years ago

PLEASE HELP

Ksenya-84 [330]

B) less
As if the atom has more electrons than protons it is negatively charged

8 0

3 years ago

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