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sasho [114]
3 years ago
13

The diameter of a hydrogen atom is approximately 1.06 ✕ 10−10 m, as defined by the diameter of the spherical electron cloud arou

nd the nucleus. The hydrogen nucleus has a diameter of approximately 2.40 ✕ 10−15 m. (a) Imagine a scale model of a hydrogen atom where its diameter is represented by the height of the Empire State Building (1454 ft to the top of its antenna). In this model, what would be the diameter of the nucleus in millimeters? mm (b) The atom is how many times larger in volume than its nucleus? Va Vn =
Chemistry
1 answer:
mylen [45]3 years ago
6 0

Answer:

a) 10.0 mm

b) 8.7 x 10¹³ times

Explanation:

Atom diameter = 1.06 x 10⁻¹⁰ m  ________________ 100%

Nucleus diameter = 2.40 x 10⁻¹⁵ m ______________   x

x = 2.26 x 10⁻³ %

The nucleus diameter is equivalent to 2.26 x 10⁻³ % of the total atom size.

a) The Empire State Building model:

1 ft = 304.8 mm

1454 ft = 443179.2 mm

443179.2 mm______ 100%

        y           ______   2.26 x 10⁻³ %

              y = 10.0 mm

In this model,  the diameter of the nucleus would be 10.0 mm.

b) Sphere volume: V =(4 · π · r³ )/3

V atom = (4 . π .( 0.53x10⁻¹⁰)³ )/3

V atom = 6.2 x 10⁻³¹ m³

V nucleus = (4 . π .( 1.2x10⁻¹⁵)³ )/3

V nucleus = 7.2 x 10⁻⁴⁵ m³

V atom / V nucleus = 6.2 x 10⁻³¹ m³ / 7.2 x 10⁻⁴⁵ m³

V atom / V nucleus = 8.7 x 10¹³

The atom is  times 8.7 x 10¹³ larger in volume than its nucleus.        

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2 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

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Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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and we all know matter involve solid liquid or gas

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Answer:

Answers may vary

Explanation:

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