Answer:I believe your answer would be option B) dilute. Hope this helps.
Explanation:
Photosynthesis. Easy ;) have a nice day
C (s) + 2 H2 (g) = CH4 (g)
Answer:
Row 1
![[H^+]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
![pH=-\log[H^+]=-\log[1.8\times 10^{-6}]=5.7](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B1.8%5Ctimes%2010%5E%7B-6%7D%5D%3D5.7)
pOh=14-pH=14-5.7=8.3
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, acidic
Row 2
![[OH^-]=3.6\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.6%5Ctimes%2010%5E%7B-10%7DM)
![pOH=-\log[OH^-]=-\log[3.6\times 10^{-10}]=9.4](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D%3D-%5Clog%5B3.6%5Ctimes%2010%5E%7B-10%7D%5D%3D9.4)
pH=14-pOH=14 - 9.4 = 4.6
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=2.6\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.6%5Ctimes%2010%5E%7B-5%7DM)
Hence, acidic
Row 3
pH = 8.15
![[H^+]=0.7\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.7%5Ctimes%2010%5E%7B-8%7DM)
pOH=14-pH=14 - 8.15 = 5.8
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=1.5\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-6%7DM)
Hence, basic
Row 4
pOH = 5.70
![[OH^-]=1.8\times 10^{-6}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.8%5Ctimes%2010%5E%7B-6%7DM)
pH=14-pOH=14 - 5.70 = 8.3
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.5\times 10^{-8}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.5%5Ctimes%2010%5E%7B-8%7DM)
Hence, basic
Answer:
there are 4 hydrogen so
A.the mass of Hydrogen in the reactant side of the equation above is 1×4=4 amu.
B.the mass of Hydrogen on the product side of the equation above =1×4=4 amu.
<u>Note</u><u>:</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>reactant</u><u> </u><u>=</u><u>mass</u><u> </u><u>of</u><u> </u><u>product</u><u>.</u>