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3 years ago

URGENT: What is different between an ionic bond and a covalent bond?

1 answer:

3 0

Ionic bond (electrovalent bond) is the exchange of ions between atoms of different element while covalent bond is the sharing electron between atoms

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What is the number of moles in 15.0 g AsH3?

pentagon [3]

We are given with the mass of Arsine ( $AsH_{3}$

The mass of arsine is 15g

there is a relation between moles, mass and molar mass of any compound which is

$moles=\frac{mass}{molarmass}$

The molar mass of Arsine = atomic mass of As + 3X atomic mass of H

the molar mass of Arsine = 74.92 + 3X 1 = 77.92 g/mol

Let us calculate the moles as

$moles=\frac{15}{77.92}=0.192mol$

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3 years ago

According to Le Chatelier’s principle, a change in pressure affects the chemical equilibrium of the reaction system under what c

charle [14.2K]

Some of the reactants or the products are in the gaseous phase.

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3 years ago

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What are the 3 boundaries of discontinuity of the earths interior

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Core , mantle, and crust

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3 years ago

What element is used in bright flashing advertising signs

Sergeu [11.5K]

The element used in advertising signs is neon.

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3 years ago

The density of thorium, which has the FCC structure, is 11.72 g/cm3. The atomic weight of thorium is 232 g/mol. Calculate (a) th

seropon [69]

Answer:

(a) a = 5.08x10⁻⁸ cm

(b) r = 179.6 pm

Explanation:

(a) The lattice parameter "a" can be calculated using the following equation:

$\rho = \frac{(N atoms/cell)*m}{V_{c}*N_{A}}$

where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and $N_{A}$ : is the Avogadro constant = 6.023x10²³ atoms/mol.

Hence the lattice parameter is:

$a^{3} = \frac{(N atoms/cell)*m}{\rho *N_{A}} = \frac{4 atoms*232 g/mol}{11.72 g/cm^{3} *6.023 \cdot 10^{23} atoms/mol} = 1.32 \cdot 10^{-22} cm^{3}$

$a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm$

(b) We know that the lattice parameter of a FCC structure is:

$a = \frac{4r}{\sqrt{2}}$

where r: is the atomic radius of Th

Hence, the atomic radius of Th is:

$r = \frac{a*\sqrt{2}}{4} = \frac{5.08 \cdot 10^{-8} cm*\sqrt{2}}{4} = 1.796 \cdot 10^{-8} cm = 179.6 pm$

I hope it helps you!

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3 years ago

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