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vova2212 [387]
3 years ago
11

A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz

. Identify the ion.
Chemistry
1 answer:
Whitepunk [10]3 years ago
3 0

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m

Using Rydberg's Formula:

\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where:

R is Rydberg constant=1.097*10^7

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99

z≅3

Atomic number is 3, So ion is Lithium ion (Li^+)

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Answer:

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Explanation:

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How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20​
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<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:  

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Reaction

Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O

  • 1. gives a coefficient

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O

  • 2. make an equation

Zn : left = a, right =1 ⇒a=1

H : left = b, right = 2d⇒ b=2d (eq 1)

N : left = b, right = 2+c⇒b=2+c (eq 2)

O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)

  • From eq 1 and eq 3

3(2d)=6+c+d

6d=6+c+d

5d=6+c (eq 4)

  • From eq 2 and eq 3

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6+3c=6+c+d

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  • From eq 4 and eq 5

5(2c)=6+c

10c=6+c

9c=6

c = 2/3

  • input eq 5

d = 2 x 2/3

d = 4/3

  • input eq 1

b = 2 x 4/3

b = 8/3

The equation

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to

Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3

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