What is the stable ion of tellurium

Chemistry

1 answer:

ivanzaharov [21]3 years ago

3 0

Answer:

Te2-

Explanation:

Tellurium, having the atomic number 52, is present in the 16th group of the periodic table. By gaining two electrons it resembles with the nearest noble gas that is Xe having the atomic number 54 . Therefore the stable ion of Tellurium is Te2-

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2 points) Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupie

Usimov [2.4K]

Answer: chlorine gas

Explanation:

mass = 0.508 g volume = 0.175 L

From ideal gas equation.

PV = nRT

(1.0 atm) × (0.175 L) = n × [0.0821 atm L / (mol K)] × [(273.2 + 25.0) K]

n = 1.0 × 0.175 / (0.0821 × 298.2) mol

n = 0.00715 mol

Molarmass= mass/mole

Molar mass of the gas = (0.508 g) / (0.00715 mol) = 71.0 g/mol

The molar mass of Cl₂ (chorine) = 35.5 × 2 g/mol = 71.0 g/mol

Hence, the gas is chlorine.

7 0

3 years ago

Find the ratio v/cv/c for an electron in the first excited state (n = 2) of hydrogen.

dezoksy [38]

The answer is 0.365:100.

v/c ratio represents ratio of speed of an electron (v) to the speed of light (c).

How is the speed of an electron calculated?

The speed of an electron (v) is given by Bohr's model as-

$v =\frac{1}{n}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h}$

Now, for the first excited state, n = 2.

e - Charge of electron = $1.6$ × $10^{-19}$ C

h - Plank's constant = $6.6$ × $10^{-34} J.s$

ε₀- permittivity $= 8.85$ × $10^{-12}$ $N^{-1}.C^{2}.m^{-2}$

Put the above data in the formula-

$v =\frac{1}{2}\; \frac{e^{2}}{4\pi \varepsilon _{0}}\times \frac{2\pi }{h} =\frac{1}{4}\times \frac{(1.6\times 10^{-19})^{2}}{8.85\times 10^{-12}\times 6.6\times 10^{-34}}\\ \\ =0.01096\times 10^{8} \\ \\=1.096\times 10^{6}ms^{-1}$