Scene B depicts chemical change in matter at atomic change.
Composition distinguishes a chemical reaction from a physical reaction. In a chemical process, the makeup of the components changes; in a physical change, the appearance, smell, or straightforward exhibition of a sample of matter changes without changing its composition. Despite the fact that we refer to them as physical "reactions," nothing is actually changing. A change in the substance in question's elemental composition is necessary for a reaction to occur. Therefore, from now on, we will simply refer to bodily "reactions" as physical changes.
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Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T = 
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where, 
= concentration =
Hence, putting the given values into the above equation as follows.
= 
= 
= 
m
or, = 
= 1 nm (approx)
Also, it is known that = 
Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.
Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
A triple bond is a type of covalent bond. It involves a sigma bond and two pi bonds (each bond involves 2 electrons). A carbon atom is capable of 4 bonds...a carbon-carbon triple bond is found in all alkynes.
Answer:
C. It does not participate in a decay series.
Explanation:
From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.
- It could have emitted any form of radioactive particles which can be alpha or beta.
- We do not know if it has a long or short half life because the value is not given.
- But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
- A decay series involves a radioactive decay in multiple steps.
