Cuales son los numeros cuanticos con ejemplo

1 answer:

4 0

Puede tomar los valores enteros positivos: n= 1, 2, 3, 4, 5, 6, 7. Por ejemplo si tengo un elemento químico que su último nivel es el 3s, su número cuántico principal sería el 3. Si tengo un elemento químico en que su último nivel es el 1s, entonces su número cuántico principal sería 1.

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If you stacked Avogadro's number of pennies one on top of the other on Earth's surface, how far would the stack extend (in kilom

attashe74 [19]

Answer:

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

Note: The question is missing some parts. The complete question is as follows;

A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]

Explanation:

Avogadro number = 6.02 * 10²³

thickness of a penny = 1.0 mm

I mm = 0.001 m

Thickness of Avogadro number of pennies stacked one upon another will be:

6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m

Distance in km;

1 m = 0.001 km

therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

5 0

3 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

Answer: The van't Hoff factor for potassium bromide in X is 1.63

Explanation:

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

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$