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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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Answer:

hco I believe

Explanation:

Pretty sure about this, when do you need to put it in?

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How many grams of H2 are needed to react with 2.40 g of N2?
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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

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A small hole in the wing of a space shuttle requires a 20.7-cm² patch.
cricket20 [7]

The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²

Given, that a space shuttle requires a 20.7 cm² patch

We have to convert the patch's area from cm² into km².

Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.

Patch area of the space shuttle is 20.7 cm²

1 cm = 0.00001 km

or, 1 cm² = (0.00001 km)²

or,  1 cm² = 10⁻¹⁰km²

20.7 cm² = 20.7 ×  10⁻¹⁰km²

20.7 cm² = 2.07 × 10⁻⁹ km²

The patch area in square kilometers is 2.07 × 10⁻⁹ km²

To learn more about unit conversion, visit: brainly.com/question/11543684

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