True or false most meteoroids burn up the ionosphere

4. Consider the following statement made in Chapter 4 of the text: "... consider that if the

AveGali [126]

The claim: "If the nucleus were the size of a grape, the electrons would be one mile away on average" is reasonably accurate because the ratios between the nucleus's sizes and the distances (between electrons and nucleus) for the two given examples are in the same order of magnitude.

To know if the claim is accurate we need to calculate the ratio of the size of the nucleus (the same as a grape) and the distance between the electrons and the nucleus for example 1 (r₁):

$r_{1} = \frac{s_{1}}{d_{1}}$ (1)

and to compare it with the ratio of the size and the distance given in example 2 (r₂):

$r_{2} = \frac{s_{2}}{d_{2}}$ (2)

Where:

s₁: is the size of the nucleus (like the size of a grape)

d₁: is the distance between electrons and nucleus of example 1 = 1 mile

s₂: is the average diameter of the nucleus = 10⁻¹³ cm

d₂: is the average distance between electrons and nucleus of example 2 = 10⁻⁸ cm

Assuming that the diameter of a grape is 3 cm (in a spherical way), the ratio of the first example is (eq 1):

$r_{1} = \frac{3 cm}{1 mi*\frac{160934 cm}{1 mi}} = 1.86 \cdot 10^{-5}$

Now, the ratio of the second example is (eq 2):

$r_{2} = \frac{10^{-13} cm}{10^{-8} cm} = 1.00 \cdot 10^{-5}$

Since r₁ and r₂ are in the same order of magnitude (10⁻⁵), we can conclude that the given claim is reasonably accurate.

You can learn more about the nucleus of an atom here: brainly.com/question/10658589?referrer=searchResults

I hope it helps you!

3 0

3 years ago

A solution of fructose, C6H12O6, a sugar found in many fruits, is made by dissolving 34.0 g of fructose in 1.00 kg of water. Wha

Mashcka [7]

Answer:

Molality → 0.188 m

Mole fraction of fructose → 0.00337

Mass percent of fructose in solution → 3.29 %

Molarity → 0.183 M

Explanation:

Solute → 34 g of fructose

Solvent → 1000 g of water

Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.

We take account density to calculate, the solution's density

1.0078 g/mL = 1034 g / mL

1034 g / 1.0078 g/mL = 1026 mL

Molal concentration → moles of solute in 1kg of solvent

Moles of fructose = mass of fructose / molar mass

34 g/ 180g/mol = 0.188 mol

0.188 mol/1kg = 0.188 m

Mole fraction of fructose = Moles of fructose / Total moles

We determine the moles of water

Moles of water = 1000 g / 18 g = 55.5 mol

Total moles = moles of fructose + moles of water

0.188 mol + 55.5 mol = 55.743 mol

0.188 mol / 55.743 mol = 0.00337

Mass percent = mass of fructose in 100 g of solution

(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %

Molarity = Moles of solute in 1L of solution

We can also say mmol of solute in 1 mL of solution

0.188 mol of fructose = 188 mmol of fructose

Molarity = 188 mmol / 1026 mL of solution = 0.183 M

8 0

3 years ago

A 25.0 ml sample of an unknown hbr solution is titrated with 0.100 m naoh. the equivalence point is reached upon the addition of

Alja [10]

1) Neutralization equation:

HBr + NaOH --> NaBr + H2O

2) 1: 1 molar ratio => 1 mol NaOH neutralizez 1 mol HBr

3) M = n / V => n = M*V

=> M*V for NaOH = M*V for HBr

0.01888 l * 0.100M = x * 0.025 l => 0.01888 l * 0.10 M / 0.025 l = 0.0755 M

Answer: 0.0755 M

3 0

3 years ago

What is a food chain

kirza4 [7]

Answer:

it is usually a restaurant that is located all over a certain country.

6 0

3 years ago

Read 2 more answers

An FM radio station broadcast at a frequency of 98.5 MHz. What is the wavelength of the station’s broadcast signal?

Troyanec [42]

Answer:

(1 MHz = 106 s -1)

Explanation:

4 0

3 years ago