The half-life of the carbon-14 isotope is used in dating fossils in a process called radiocarbon dating.
hope this is adequate.
Answer:
pH of 7.86, then the [OH-] is equal to 10^(14-7.86) = 10^6.14 M
Explanation:
The empirical formula is a formula of a compound showing the proportion of each element involved in the compounds but it does not represent the total number of atoms in the compound. It is the lowest number of ratio between the elements in the compound. In order, to determine the actual number of the atoms or the molecular formula of the compounds, we make use of the molar mass of the compound.
<span>To
determine the molecular formula, we multiply a value to the empirical formula.
Then, calculate the molar mass and see whether it is equal to the one
given (104.1 g/ mol). From the choices, the only valid options are b, d and e.
</span> molar mass
1 CH 13.02
8 C8H8 104.16
6 C6H6 78.12
Therefore the correct answer is option B.
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺
|| Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu
⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺
+ 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu
⇄ Cu²⁺
+ 2e⁻
Ag⁺
+ e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺
+ 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu
+ 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu
+ 2Ag⁺
⇄ Cu²⁺
+ 2Ag
20 g O2 x 1 mol O2/32 g O = 0.625 mol O2