The mass of substance which consists the equal number of particles as 12 grams of Carbon-12 is said to be mole.
For example:
The balanced chemical equation for production of water from hydrogen and oxygen is:
That means 2 moles of reacts with 1 mole of to give 2 moles of .
The given formula is .
Hence, 4 represents the number of moles in formula that is 4 moles of .
The number of atoms of sodium, in is:
The number of atoms of sulfur, in is:
The number of atoms of oxygen in, in is:
Answer:
The names of the electron shell were given by a spectroscopist named Charles G Barkla. He named the innermost shell has k shell because he noticed that the X-rays emitted two types energies. These energies were named as type A that is higher energy X-ray and type B that is lower energy X-ray. Later these energiers were named with different alphabets. He noticed that K type X-rays emitted the highest energy. Therefore, he named the innermost shell as the K shell.
Explanation:
<h2>#hope</h2>
the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
<h3 />
FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
Learn more about % yield here:brainly.com/question/27979178
#SPJ10
<u>Answer: </u>
The pressure of a sample of argon gas was increased from 2.12atm to 6.96atm at constant temperature. If the final volume of the argon sample was 16.9L
. The initial volume of the argon sample was 55.5L
<u>Explanation:</u>
Considering the initial volume, initial pressure to be V1 and P1 respectively
Assuming the final volume and final temperature of the gas to be V2 and P2 respectively
Given,
Pressure of the gas was increased from 2.12 atm to 6.96 atm
Therefore, P1 = 2.12 atm
P2 = 6.96 atm
Final volume, V2 = 16.9 L
Applying Boyle’s law,
Substituting the value
V1 = 55.5 L
Therefore, the initial volume was 55.5L