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wolverine [178]
3 years ago
11

What time did frosty die in frosty meets his demise

Chemistry
1 answer:
sergejj [24]3 years ago
8 0

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:

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Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi
uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

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