d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
<h3>magma that is rich in feldspar and silica is called STRATOVOLCANO.</h3>
Answer: the correct answer would be the last option, C6H12O6 and O2
Explanation:
The melting and boiling point depend on the strength of the ihydrogen bonds. Hydrogen bonding will cause the higher the melting and boiling points because more energy is needed to break bonds between molecules.
Hydrogen bonds affect solubility in water, molecules with hydrogen bonds dissolve better in water.
Answer:
- <u>2.59 × 10⁻⁷ m = 259 nm</u>
Explanation:
You need to calculate the wavelength of a photon with an energy equal to 463 kJ/mol, which is the energy to break an oxygen-hydrogen atom.
The energy of a photon and its wavelength are related by the Planck - Einstein equation:
Where:
- h = Planck constant (6.626 × 10⁻³⁴ J . s) and
- ν = frequency of the photon.
And:
Where:
- c = speed of light (3.00 × 10⁸ m/s in vacuum)
- λ = wavelength of the photon
Thus, you can derive:
Solve for λ:
Before substituting the values, convert the energy, 463 kJ/ mol, to J/bond
- 463 kJ/ mol × 1,000 J/kJ × 1 mol / 6.022 × 10 ²³ atom × 1 bond / atom
= 7.69×10²³ J / bond
Substitute the values and use the energy of one bond:
- λ = 6.626 × 10⁻³⁴ J . s × 3.00 × 10⁸ m/s / 7.69×10²³ J = 2.59 × 10⁻⁷ m
The wavelength of light is usually shown in nanometers:
- 2.59 × 10⁻⁷ m × 10⁹ nm / m = 259 nm ← answer
