<h3>
Answer:</h3>
1.43 × 10⁻²⁰ mol Li
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
8.63 × 10³ atoms Li
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
Answer:
correct option is A) atmospheric pollutants
Explanation:
- Nuclear power plant do not produce air pollution in the form of carbon dioxide and sulfur dioxide.
- Nuclear as radioactive waste is hazardous to all kind of life and the environment.
- Long-term storage of radioactive waste is required. It is important that waste stabilization is a form that does not react or deteriorate for years.
- If nuclear waste is wasted in the atmosphere or in the oceans and lagre lakes, it eventually comes into contact with organisms.
Answer:
174 kPa
Explanation:
Given that,
Initial temperature, T₁ = 25° C = 25+273 = 298 K
Final temperature, T₂ = 225°C = 225 + 273 = 498 K
Initial pressure, P₁ = 104 kPa
We need to find the new pressure. The relation between the temperature and pressure is given by :
So,
or
P₂ = 174 kPa
So, the new pressure is 174 kPa.
