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lara [203]
2 years ago
11

The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

it is a sphere).
Chemistry
2 answers:
Sladkaya [172]2 years ago
4 0

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = 1  \ atom \times  (\dfrac{1  \ mole}{  6.022 \times  10^{23}  \ atoms}) \times ( \dfrac{4.003 \ grams}{  1  \ mole})

=6.647 \times  10^{-24} \  grams

The volume can be determined as  folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = \dfrac{0.10}{2} = 0.05 nm

Converting it into cm, we have:

0.05 nm \times  \dfrac{10^{-9} \  meters}{ 1  nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}

=5 \times  10^{-9}  \ cm

Assuming that it is a sphere, the volume of a sphere is

= \dfrac{4}{3}\pi r^3

= \dfrac{4}{3}\pi  \times (5\times 10^{-9})^3

= 5.236 \times 10^{-25} \ cm^3

Finally, the density can be calcuated by using the formula :

Density = \dfrac{mass}{volume}

D =  \dfrac{6.647 \times  10^{-24} \  grams }{ 5.236 \times 10^{-25} \  cm^3}

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

Damm [24]2 years ago
3 0

Answer:

<h2><em> 12.7g/cm³</em></h2>

Explanation:

Density is defined as the ratio of mass per unit volume of a substance.

Density = Mass/Volume

Since the shape of the Helium atom is a sphere, we will use the formula for calculating the volume of a sphere to get the volume of helium

Volume of a sphere = 4/3πr³ where r is the radius of the sphere.

Given the diameter of the sphere = 0.10nm

radius of the sphere = 0.10/2 = 0.05nm = 5*10⁻⁹cm

Volume of the sphere = 4/3 * 3.14 * (5*10⁻⁹)³

V =  4/3 * 3.14 * 125*10⁻²⁷

V = 1,570*10⁻²⁷/3

V = 523.33*10⁻²⁷

V =  5.233*10⁻²⁵cm³

Mass of the Helium atom = 6.6464731 × 10⁻²⁴ g

Density of the Helium atom in g/cm³ = 6.6464731 × 10⁻²⁴ g/5.233*10⁻²⁵cm³

Density of the Helium atom in g/cm³ = 1.27 * 10⁻²⁴⁻⁽⁻²⁵⁾g/cm³

Density of the Helium atom in g/cm³ = 1.27 * 10⁻²⁴⁺²⁵g/cm³

Density of the Helium atom in g/cm³ = 1.27 * 10¹g/cm³

Density of the Helium atom in g/cm³ = 12.7g/cm³

<em>Hence the density of the He atom in g/cm³ is 12.7g/cm³</em>

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mamaluj [8]

Answer:

an amine group

Explanation:

Hope that helped :)

5 0
2 years ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)
Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0
2 years ago
The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have
Dmitrij [34]

Explanation:

The highest principal energy level of period 2 elements is 2. (True)

The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

Period 3 elements all have six 3p electrons – (False)  

For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)  

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

The highest principal energy level of period 3 elements is 4 – (False)

The highest principle energy level of elements of any period is equal to period number.

For example, sodium is present in 3rd period, so its principle quantum number will be 3.

Period 5 elements have an inner electron configuration of [Xe] – (False)

Period 5 elements have an inner electron configuration of [Kr]. The first member of this period is rubidium. 37 and that of Kr is 36.  

6 0
2 years ago
You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl
salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

7 0
2 years ago
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