The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
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What is Mole ?</h3>
A mole is a very important unit of measurement that chemists use.
A mole of something means you have 6.023 x 10 ²³ of that thing.
- For 2.15 mol of hydrogen sulphide (H₂S) :
1 mole hydrogen sulphide (H₂S) = 34.08088 grams
Therefore,
2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol
= 73.272 gm
- For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;
1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams
Therefore,
3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol
= 1.82 gm
Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
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Answer:
No
Explanation:
The conclusions from Thomoson's claims would be invalid if his experiment could not be replicated.
<em>One of the attributes of a valid experiment in science is replication. That is, an experiment must be repeatable with similar outcomes under similar conditions as the original experiment when independently performed by another scientist. Once an experiment cannot be replicated, the outcomes of such an experiment become invalid in science.</em>
Hence, <u>Thomoson's conclusion would be invalid if his experiment could not be replicated. </u>
Answer:
12.7mol Na.
Explanation:
Hello there!
In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

Thus, by performing the division we obtain:

Regards!
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
Yeah no one is gonna read all that babe.