The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.
The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.
The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.
Thorium has several isotopes, with different rates and types of emission:
- ²³²Th emits α particles, it is the most abundant 99.9%
- ²³⁴Th emits β particles, exists in small traces.
In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.
Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.
In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:
-
On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.
- The neutrons cannot be detected in this experiment because they have no electrical charge.
Learn more about radioactive emission here: brainly.com/question/15176980
The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =
The empirical formula : MnO₂
Answer: Mixtures do not have the same composition, properties, and appearance throughout.
Explanation: It does not vary from sample to sample.
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
