Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is
and written as OH.
Answer:
oxidation- reduction
Explanation:
where gaining electronic reduces one element and losing them oxidize the other nitric acid is not only strong it is also a oxidizing agent
<h2>Oxidize: copper = Cu+2</h2>
Explanation:
The given data is as follows.
Thickness (dx) = 0.87 m, thermal conductivity (k) = 13 W/m-K
Surface area (A) = 5 ,
According to Fourier's law,
Q =
Hence, putting the given values into the above formula as follows.
Q =
=
= 5902.298 W
Therefore, we can conclude that the rate of heat transfer is 5902.298 W.
<u>Answer and Explanation:</u>
Mercury combines with sulfur as follows -
Hg + S = HgS
Hg = 200,59
S = 32,066 Therefore 1.58 g of Hg will react with -
1.58 multiply with 32,066 divide by 200,96 of sulfur.
= 0.25211 g S
This will form 1.58 + 0.25211 g HgS = 1.83211 g HgS
The amount of S remaining = 1.10 - 0.25211 = 0.84789 g