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gizmo_the_mogwai [7]
3 years ago
10

The process n01 → p11 + β−10 n 0 1 → p 1 1 + β - 1 0 represents a(n) _____.

Chemistry
1 answer:
polet [3.4K]3 years ago
5 0

Answer:

  • <u><em>beta decay</em></u>

Explanation:

The <em>process</em> is represented by the nuclear equation:

  • ^1_0n → ^1_1p+^0_{-1}\beta

Where:

  • n represents a neutron,
  • p represents a proton, and
  • β represents an electron.

The superscripts to the leff of each symbol is the mass number (number of protons and neutrons), and the subscript to the left means the atomic number (number of protons).

Then, in this process a neutron is being transformed into a proton by the emssion of an electron (from inside the nucleus of the atom).

This electron is named beta (β) particle, and the process is called <u><em>beta decay</em></u>, because the neutron is changing into other subatomic particles.

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For at least the last century Florida has experienced a hurricane season every year, from June to November. Which of the followi
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Answer:

Florida's Global Pattern.

Explanation:

Because of where Florida sits, the weather in the Gulf of Mexico actually blows enough water to Create tidal waves, and also hurricanes. at least that's what I was told cause I live here :>

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Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.
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Answer:

Explanation:

 The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu  has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so  Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

7 0
3 years ago
At 500 K the reaction 2 NO(g) + Cl2(g) ⇌ 2 NOCl(g) has Kp = 51 In an equilibrium mixture at 500 K, the partial pressure of NO is
Aleks [24]

Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }

Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)

7 0
2 years ago
Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
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