Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.
Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.
Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Learn more about electrode here:
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Answer:3.6 I think sorry if wrong
Explanation:
90 divided by 25
Answer:
5.41 ×10⁻²²
Explanation:
We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.
Then, the number of ZnO molecule in one unit cell = 4
The standard molar mass of ZnO = 81.38g
Avogadro's constant = 6.023 × 10²³ mole
∴
The mass of one unit cell of zinc oxide can be calculated as:
= 
= 5.40461564×10⁻²²
≅ 5.41 ×10⁻²²
∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²
