Answer:
No photoelectric effect is observed for Mercury.
Explanation:
From E= hf
h= Plank's constant
f= frequency of incident light
Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23
f= 11×10^14 Hz
The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface
(1) Ocean to Continent
(2)Continent to Continent
(3)Ocean to Ocean
are the three sub types of convergent plate boundaries.
At the boiling point, the vapor pressure of a liquid is equal to the atmospheric pressure.
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
Answer:
The symbol of the ion is A^-
Explanation:
Let A be the symbol of the element.
Proton = 17
Electron = 18
Neutron = 20
Since the element has more electrons than protons, it means it has gain electrons.
From the question given,
The difference between the electron and proton = 18 — 17 = 1
So, the element has gain 1 electron.
Therefore the symbol of the ion is A^-
