Step one: Identify reactants and products and place them in a word equation.
Step two: Convert the chemical names into chemical formulas. Place them based on the chemical equation and write the state symbols.
Step three: balance the chemical equation.
Answer: false
Explanation:
It is false that to study Earth's interior, geologists often rely on indirect methods, such as evidence from fossils. They rely on seismic wave.
Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
