The Halogen that is most likely easy to be oxidised would be the Iodine. In addition, the property that makes the Iodine ideal for oxidation is due to the fact that its d-orbital contains more electrons compared to other halogens such as fluorine where oxidation would tend to be difficult to occur on the element.
Answer:
2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20
Explanation:
Since there's no "?" shown in the equation, let's balance it and solve it entirely.
Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20
1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.
To do this, let's start by looking at the elements that are only presnet once on each side:
On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H → to balance it, you need 4 water molecules on the products side.
Answer:
4.36 ppm
Explanation:
First we convert percent transmittance to absorbance:
Then we <u>calculate the concentration of the solution</u>, using <em>Lambert-Beer's equation</em>:
It is usually used with molar concentrations but given that the given absortivity is in ppm terms and the answer is also in ppm, we can simply use the given value.
- 0.268 = 6.14x10⁻²L·ppm⁻¹ * 1.00 cm * C
The answer is 64.907 amu.
The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2
We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083
Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
