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2 years ago

How does an atom of tellerium become an ion.

1 answer:

8 0

Answer: Element Tellurium (Te), Group 16, Atomic Number 52, p-block, Mass 127.60. ... defined as being the charge that an atom would have if all bonds were ionic.

What ion does tellurium form?

Te+4

Tellurium, ion (Te4+) | Te+4 - PubChem.

Explanation: Tellurium is a chemical element with the symbol Te and atomic number 52. It is a brittle, mildly toxic, rare, silver-white metalloid.

Tellurium

Tellurium Element - Visual Elements Periodic Table

Discovery date 1783

Discovered by Franz-Joseph Müller von Reichenstein

Origin of the name The name is derived from the Latin 'tellus', meaning Earth.

Allotropes

Tellurium

52127.60

Fact box

Group 16 Melting point 449.51°C, 841.12°F, 722.66 K

Period 5 Boiling point 988°C, 1810°F, 1261 K

Block p Density (g cm−3) 6.232

Atomic number 52 Relative atomic mass 127.60

State at 20°C Solid Key isotopes 130Te

Electron configuration [Kr] 4d105s25p4 CAS number 13494-80-9

ChemSpider ID 4885717 ChemSpider is a free chemical structure database

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For the reaction, X + Y → A + B, ΔGo is –1324 kJ. Which one of the following statements is NOT valid concerning the reaction?

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2 years ago

What affect will adding sugar have on the viscosity of water?

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3 years ago

Read 2 more answers

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

In an aqueous solution, 42% of a substance dissociates to release hydronium ions. Which of the following statements is true for

const2013 [10]

Answer: The statement it is a weak acid is true for the substance.

Explanation:

An acid that dissociates completely when dissolved in water to give hydrogen $(H^{+})$ or hydronium $(H_{3}O^{+})$ ions is called a strong acid.

For example, HCl is a strong acid.

$HCl + H_{2}O \rightarrow H_{3}O^{+} + Cl^{-}$

An acid that dissociates partially or weakly when dissolved in water to given hydrogen or hydronium ions is called a weak acid.

For example, $CH_{3}COOH$ is a weak acid.

$CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H_{3}O^{+}$

A strong base is a base which when dissolved in water then it dissociates completely to give hydroxide ions.

For example, NaOH is a strong acid.

A weak base is a base which when dissolved in water then it dissociates partially or weakly to give hydroxide ions.

For example, $NH_{3}$ is a weak base.

Hence, in an aqueous solution where 42% of a substance dissociates to release hydronium ions shows that the dissociation is less than 50%. This means that substance is dissociating weakly so, it is a weak acid.

Thus, we can conclude that the statement it is a weak acid is true for the substance.

5 0

2 years ago