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
Answer:
Demo Mole Quantities
58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na
+
Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)
63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu
19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al
Explanation:
The answer is zero!
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The component of the dri reflects the requirement of a nutrient for 50 percent of healthy americans and canadians in a specific life stage and gender is Estimated Average Requirement (EAR).
What is EAR?
A daily food intake number known as the Estimated Average Requirement (EAR) is one that is thought to satisfy the needs of 50% of healthy persons in a given life stage and gender group.
Based on a review of the scientific literature, the estimated average requirements (EAR) are predicted to meet the needs of 50% of the population in that age range.
It serves as the foundation for the creation of RDAs and is employed to assess the sufficiency of nutritional consumption for a given population.
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The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
To view more about rational reaction, refer to:
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