The first one is abundant
Answer: 4.74 moles
Explanation:
Moles of Ba=
Moles of Ba=
According to given balanced equation,
1 mole of Barium gives 2 moles of Li
Thus 2.37 moles of Barium gives=
Thus 4.74 moles of lithium will be produced by 135 g of barium.
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
Mendeleev positioned elements in the periodic table in increasing order of their atomic numbers, such that elements having identical chemical properties and characteristics plunge into the same group.
<h3>What is Mendeleev's periodic table?</h3>
Mendeleev's periodic table may be defined as a collection of elements in an increasing atomic mass hierarchy in a table arrangement, such that it reminisces resemblances and manias according to their chemical properties and characteristics.
Mendeleev found that two elements existed between atomic weights 65.2 and 75 because he comprehended especially pleasingly that the possessions of the elements were more comparable and closer to this degree.
He also anticipated having other elements that would have their possessions comparable to these other elements.
Therefore, he departed a void for these two elements in the periodic table until they were ultimately found in their real existence.
The complete question is as follows:
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known with atomic weights between 65.2 and 75. But Mendeleev predicted that two elements must exist with atomic weights in this range. What led Mendeleev to predict that two undiscovered elements existed in that range?
Therefore, it is well described above.
To learn more about Mendeleev's periodic table, refer to the link:
brainly.com/question/13991055
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Answer:
M₂ = 0.23 M
Explanation:
Given data:
Volume of HNO₃ = 22.5 mL
Volume of NaOH = 31.27 mL
Molarity of NaOH = 0.167 M
Molarity of HNO₃ = ?
Solution:
M₁ = Molarity of NaOH
V₁ = Volume of NaOH
M₂ = Molarity of HNO₃
V₂ = Volume of HNO₃
M₁V₁ = M₂V₂
0.167 M × 31.27 mL = M₂ × 22.5 mL
5.2 M. mL = M₂ × 22.5 mL
M₂ = 5.2 M. mL /22.5 mL
M₂ = 0.23 M
