Answer:
this lesson is the third in a three-part series about the nucleus, isotopes, and radioactive decay. The first lesson, Isotopes of Pennies, deals with isotopes and atomic mass. The second lesson, Radioactive Decay: A Sweet Simulation of Half-life, introduces the idea of half-life.
By the end of the 8th grade, students should know that all matter is made up of atoms, which are far too small to see directly through a microscope. They should also understand that the atoms of any element are alike but are different from atoms of other elements. Atoms may stick together in well-defined molecules or they could be packed together in large arrays.
For students, understanding the general architecture of the atom and the roles played by the main constituents of the atom in determining the properties of materials now becomes relevant. Having learned earlier that all the atoms of an element are identical and are different from those of all other elements, students now come up against the idea that, on the contrary, atoms of the same element can differ in important ways. (Benchmarks for Science Literacy, p. 79.)
In this lesson, students will be asked to consider the case of when Frosty the Snowman met his demise (began to melt). The exercise they will go through of working backwards from measurements to age should help them understand how scientists use carbon dating to try to determine the age of fossils and other materials. To be able to do this lesson and understand the idea of half-life, students should understand ratios and the multiplication of fractions, and be somewhat comfortable with probability
Explanation:
Answer:
The total pressure after one half is 6.375 atm.
Explanation:
The initial pressure of product is increases while the pressure of reactant would decrease.
Balanced chemical equation:
2N₂O → 2N₂ + O₂
The pressure of N₂O is 5.10 atm. The change in pressure would be,
N₂O = -2x
N₂ = +2x
O₂ = +x
The total pressure will be
P(total) = P(N₂O) + P(N₂) + P(O₂)
P(total) = ( 5.10 - 2x) + (2x) + (x)
P(total) = 5.10 + x
After one half life:
P(N₂O) = 1/2(5.10) = 5.10 - 2x
x = 5.10 - 1/2(5.10) /2
x = 5.10 - 0.5 (5.10) /2
x = 5.10 - 2.55 / 2
x = 2.55 /2 = 1.275 atm
Thus the total pressure will be,
P(total) = 5.10 + x
P(total) = 5.10 + 1.275
P(total) = 6.375 atm
If a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, then the final volume would be 11,943.8144 ml if the pressure of the gas is changed to 775 torr assuming that the amount and the temperature of the gas remain constant.
It is given that the initial pressure P₁ is 24,650Pa and initial volumeV₁ is 376ml and the final pressureP₂ is 775 torr. We need to find the final volume of the gas. The final volume could be found using the following formula:
P₁V₁ = P₂V₂
By substituting the values, we get
24650 x 376 = 776 x V₂
9268400 = 776V₂
V₂ = 9268400/776
V₂ = 11,943.8144 ml
Therefore, the final volume of the gas would be 11,943.8144 ml
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Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
