Upper N upper H subscript 4 upper C l (s) right and left arrows stacked above each other upper N upper H subscript 3 (g) plus upper H upper C l (g)
Explanation:
The given equation is;
NH₄Cl ⇄ NH₃ + HCl
This equation is clearly different from the other ones.
- In the reactant going forward, there is a right and left arrows stacked above each other.
- The symbol is ⇄ and it is used to show reversibility of chemical reactions.
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Answer:
a: 1st paragraph from the left side indicates Newton's first law.
b:, 2nd paragraph from the left side indicates Newton's third law.
c: 1st paragraph from right side indicates Newton's first law.
d: 2nd paragraph from right side indicates Newton's second law.
Explanation:
Answer:
Explanation:
You look at the type of atom and their electronegativity difference.
If ΔEN <1.6, covalent; if ΔEN >1.6, ionic
Ar/Xe: Noble gases; no reaction
F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic
N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.
Exothermic reaction releases energy in the form of heat or light
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
