B) 40%
The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.
1. When our Galaxy was forming why was the center dense?
Answer: <u>Stars are believed to originate as from gas clouds in space. Gravitational forces cause this gas to contract forming a dense center of gravitational force.Eventually this extreme pressure at the center of the star causes nuclear fusion reactions to occur.</u>
2. Why did hydrogen atoms collide in the dense center of the Galaxy?
Answer: <u>Compression would heat the gas to temperatures above 1,000 kelvins. Some hydrogen atoms would pair up in the dense, hot gas, creating trace amounts of molecular hydrogen. The hydrogen molecules would then start to cool the densest parts of the gas by emitting infrared radiation after they collide with hydrogen atoms.</u>
Answer : It takes time for the concentration to decrease to 0.100 M is, 22.4 s
Explanation :
Formula used to calculate the rate constant for zero order reaction.
The expression used is:
![\ln [A]=-kt+\ln [A_o]](https://tex.z-dn.net/?f=%5Cln%20%5BA%5D%3D-kt%2B%5Cln%20%5BA_o%5D)
where,
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration = 0.537 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= final concentration = 0.100 M
t = time = ?
k = rate constant = 0.075 M/s
Now put all the given values in the above expression, we get:
Therefore, it takes time for the concentration to decrease to 0.100 M is, 22.4 s
Answer:
11
Explanation:
Moles of KOH = 
Volume of water = 10 liters
Concentration of KOH is given by
![[KOH]=\dfrac{10^{-2}}{10}\\\Rightarrow [KOH]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5Cdfrac%7B10%5E%7B-2%7D%7D%7B10%7D%5C%5C%5CRightarrow%20%5BKOH%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
![[KOH]](https://tex.z-dn.net/?f=%5BKOH%5D)
is strong base so we have the following relation
![[KOH]=[OH^{-}]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E%7B-%7D%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
![pOH=-\log [OH^{-}]=-\log10^{-3}](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E%7B-%7D%5D%3D-%5Clog10%5E%7B-3%7D)
So, pH of the solution is 11
