Answer:
it's C
Explanation:
because it exhaled the carbon dioxide
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
![[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
![[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:
V = 12.5 L
Explanation:
Given data:
Volume of NO = 15.0 L
Temperature and pressure = standard
Volume of nitrogen gas produced = ?
Solution:
Chemical equation:
6NO + 4NH₃ → 5N₂ + 6 H₂O
Number of moles of NO:
PV = nRT
n = PV/RT
n = 1 atm × 15.0 L / 0.0821 atm.L /mol.K × 273.15 K
n = 15.0 atm.L / 22.43 atm.L /mol
n = 0.67 mol
now we will compare the moles of No and nitrogen gas.
NO : N₂
6 : 5
0.67 : 5/6×0.67 = 0.56
Volume of nitrogen gas:
PV = nRT
1 atm × V = 0.56 mol × 0.0821 atm.L /mol.K × 273.15 K
V = 12.5 atm.L / 1 atm
V = 12.5 L
