Answer:
V₂ = 21.3 dm³
Explanation:
Given data:
Initial volume of gas = 3.00 dm³
Initial pressure = 101 Kpa
Final pressure = 14.2 Kpa
Final volume = ?
Solution;
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
101 Kpa × 3.00 dm³ = 14.2 Kpa × V₂
V₂ = 303 Kpa. dm³/ 14.2 Kpa
V₂ = 21.3 dm³
Explanation:
Answer:
The area around the nucleus must be of low mass.
Explanation:
Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.
It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.
Hope this info is useful.
<span>we know that each
element has an unique spectra and it can be used to identify the
element. it shows that the energy levels of the electrons and different colors are the result of different wavelengths.
hope it helps
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Malleability is the property due to which substances tend to hammered into thin sheets when pressure is applied. Malleability is observed in metal as the metal atoms are bound by metallic bonds. The layers of metal atoms can roll over each other without breaking the metal bond when pressure is applied. In the periodic table malleability decreases as the metallic nature decreases. Among the given metals Al is the most metallic element, followed by Zinc an Carbon is a non metal. Therefore, the order of increasing malleability will be C, Zn, Al
Answer:
if a mixture of a given percentage or ratio strength is diluted to twice its original quantity, its active ingredient will be contained in twice as many parts of the whole, and its strength therefore will be reduced by one-half
