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Ainat [17]
3 years ago
10

Science question down below

Chemistry
1 answer:
slavikrds [6]3 years ago
4 0
A. Reproduction

Birth is producing children just as they were which is REproduction

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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
Sladkaya [172]

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

4 0
3 years ago
Read 2 more answers
Which of the following is NOT true of white dwarfs?
liq [111]

Answer:

A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. Near the end of its nuclear burning stage, this type of star expels most of its outer material, creating a planetary nebula. Only the hot core of the star remains. ... That means a white dwarf is 200,000 times as dense.

Explanation:

they are cold

they are about the size of Earth

these both are not true

6 0
3 years ago
Read 2 more answers
Question 3 of 8
vitfil [10]

Answer:

B

Explanation:

7 0
2 years ago
Read 2 more answers
What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4Cl produced by the reaction
Mashcka [7]

Answer:

a) HCl is the limiting reagent.

b) Mass of NH₄Cl formed = 6.68 g

c) Pressure of the gas remaining in the flask = 1.742 atm

Explanation:

The complete Question is presented in the attached image to this solution.

To solve this question, we first need to obtain the limiting regaent for this reaction.

The limiting reagent is the reagent that is in short supply in the reaction and is used up in the reaction. It determines the amount of products that will be formed and the amount of other reactants that will be required for the reaction.

NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)

1 mole of NH₃ reacts with 1 mole of HCl

we first convert the masses of the gases available to number of moles.

Number of moles = (Mass/Molar Mass)

Molar mass of NH₃ = 17.031 g/mol, Molar mass of HCl = 36.46 g/mol

Number of moles of NH₃ = (4.55/17.031) = 0.2672 mole

Number of moles of HCl = (4.55/36.46) = 0.1248 mole

Since 1 mole of NH₃ reacts with 1 mole of HCl

It is evident that HCl is in short supply and is the limiting reagent.

NH₃ is in excess.

So, to calculate the amount of NH₄Cl formed,

1 mole of HCl gives 1 mole of NH₄Cl

0.1248 mole of HCl will also gove 0.1248 mole of NH₄Cl

Mass (Number of moles) × (Molar Mass)

Molar mass of NH₄Cl = 53.491 g/mol

Mass of NH₄Cl formed = 0.1248 × 53.491 = 6.68 g

c) The gas remaining in the flask is NH₃

0.1248 mole of NH₃ is used up for the reaction, but 0.2672 mole was initially available for reaction,

The amount of NH₃ left in the reacting flask is then

0.2672 - 0.1248 = 0.1424 mole.

Using the ideal gas Equation, PV = nRT

We can obtain the rrequired pressure of the remaining gas in the flask

P = Pressure = ?

V = Volume = 2.00 L

n = number of moles = 0.1424 mole

R = molar gas constant = 0.08205 L.atm/mol.K

T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K

P = (nRT/V)

P = (0.1424×0.08205×298.15/2) = 1.742 atm

Hope this Helps!!!

7 0
3 years ago
Consider the reaction described by the following chemical equation.
oee [108]

Answer:

Q = -897 kJ/mol

Explanation:

From the given information:

The heat released Q = -65.9 kJ

To start with the molar mass of H_2O_2 = 2 × (molar mass of H) + 2 × (molar mass of O)

= (2 × 1.008) + (2 × 16.0 )

= 34.016 g/mol

However, given that:

mass of H_2O_2  2.50 g

The number of moles of H_2O_2  = \dfrac{mass}{molar \ mass}

= \dfrac{2.5}{34.016}

= 7.349 \times 10^{-2} \ mol

Finally; Using the formula:

\Delta H = \dfrac{Q}{number \ of \ moles}\\ \\ Q = \dfrac{-65.9 \ kJ}{7.349 \times 10^{-2} \ mol}

Q = -897 kJ/mol

6 0
2 years ago
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