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3 years ago

What generalizations can be made about silver halides?

2 answers:

3 0

Halides is the term given to the ions of halogens. Halogens are the second-to-the-last column or period in the periodic table. Examples are chlorine, fluorine, bromine and iodine. Halides are all soluble in water except when combine with silver, lead and mercury. Therefore, the generalization we can make is that silver halides are insoluble in water,

sweet-ann [11.9K]3 years ago

3 0

<h2>Generalization can be made about Silver Halides </h2>

Silver halides are the compounds which are formed by the reaction of silver and halogens. Halogens are fluorine, chlorine, bromine, iodine and sometimes astatine. So the compounds are formed such as silver fluoride, silver bromide, silver chloride and silver iodide.

Silver halides are insoluble in water except for silver fluoride (AgF). Silver chloride (AgCl) forms a white precipitate. Silver bromide (AgBr) gives a creamy coloured precipitate and silver iodide (AgI) a yellow coloured precipitate.

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MariettaO [177]

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2 years ago

Read 2 more answers

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

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2 years ago

Activity

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Answer:

The mass of a pure substance represents a specific number of pure substance 'molecules', as defined by molecular weight. The relative atomic mass (atomic weight) in grams/mole (g/mol) is the molar mass of a substance, which means it is the mass of one mole of that substance.

Explanation:

Hope this helps :)

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3 years ago

Imagine that Zapata was able to fly his machine far out into outer space. What do you predict would happen to Zapatas forward mo

scoray [572]

Answer:

B if they run out of fuel then they would probably keep going because there is nothing that would stop them other than like a planet or something.

Explanation:

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2 years ago

Which statement about fusion reactors is true? a Enriched uranium isotopes are used as fuel in fusion reactors. b Fuel for fusio

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Answer:

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Explanation:

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2 years ago