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skad [1K]
3 years ago
10

What type(s) of intermolecular forces are expected between seh2 molecules?

Chemistry
2 answers:
bija089 [108]3 years ago
7 0
London dispersion
dipole-dipole
olga_2 [115]3 years ago
3 0

Answer:

Induced and dipole forces.

Explanation:

Hello,

Due to its structure, SeH_2, one identifies the following two intermolecular forces:

- Induced: Ion-Induced Dipole Forces

At first, an ion-induced dipole attraction is present as a weak force which results when the approach of an ion induces a dipole in this nonpolar molecule by disturbing the arrangement of electrons.

- Dipole forces.

Now, dipole-dipole forces are present in such molecule as attractive forces between the positive end of one of the polar molecule and the negative end of another polar space in the molecule.

Best regards.

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When water is boiling the particles in the liquid are not expanding. This also applies for other liquids. Particles do not expand, it is only the volume they take up that expands. When water is boiling, the particles are rather escaping the liquid phase by undergoing a phase change which forms the particles into the gaseous phase.
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How many moles are in 3.57 x 10 32<br> particles Ca?
valentina_108 [34]

Answer:

\boxed {\boxed {\sf 592,826,303.6 \ moles \ of \ Ca }}

Explanation:

To convert from moles to particles, we must Avogadro's Number.

6.022*10^{23}

This number tells us the number of particles (atoms, ions, molecules, etc.) in 1 mole of a substance. In this case, it is the particles of calcium in 1 mole of calcium.

6.022*10^{23} \ particles \ Ca / mole \ Ca

We can use Avogadro's Number as a ratio or fraction.

\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}

Multiply this by the given number of particles (3.57*10³²).

3.57*10^{32} \ particles \ Ca*\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}

Flip the fraction so the particles of calcium will cancel out.

3.57*10^{32} \ particles \ Ca*\frac{1 \ mol \ Ca}{6.022 *10^{23} \ particles \ Ca}

3.57*10^{32} *\frac{1 \ mol \ Ca}{6.022 *10^{23} }

\frac{3.57*10^{32} \ mol \ Ca}{6.022 *10^{23} }

592826303.6 \ mol \ Ca

There are <u>592,826,303.6 moles</u> of calcium in 3.57*10³² particles of calcium.

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Read the given equation.
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<h2>Answer:</h2>

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