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2 years ago

Calculate the volume of material whose density is 0.798gcm-3 and a mass 25g

Physics

2 answers:

luda_lava [24]2 years ago

5 0

Explanation:

density equals mass/volume

Elena L [17]2 years ago

4 0

Answer:

$\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

Explanation:

To calculate the volume, we can use the following formula:

$\boxed{\mathrm{density = \frac{mass}{volume}}}$ .

In this case,

• density = 0.798 gcm⁻³

• mass = 25 g

Substituting values into the formula:

$\mathrm{0.798 \space\ gcm^{-3}= \frac{25 \space\ g}{volume}}$

⇒ $\mathrm{volume = \frac{25 \space\ g}{0.798 \space\ gcm^{-3}}}$

⇒ $\mathrm{volume = \bf 31.33 \sapce\ cm^{-3}}$

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1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

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Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

$F = \kappa_{0} \frac{q1q2}{r^{2}}$ (1)

Where q1 and q2 are the charges, $\kappa_{0}$ is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (for that case will be the first ionization energy).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10

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