Answer:
1.26x10^25 atoms of hydrogen
Explanation:
because there are 12 atoms of hydrogen in a molecule of glucose, multiply 12 by Avogadro's number (6.02x10^23) to get how many molecules of hydrogen there are in a mole of glucose. Then multiply that number by 1.75, which is the number of moles of glucose there is in this problem.
Safety glasses should be worn any time you are doing an experiment, especially one that involves chemicals or chemical reactions. They prevent chemicals or other materials from getting on or in your eye, and can prevent anything from mild discomfort to permanent blindness.
Some pairs of safety glasses have magnifying glasses on them, similar to bifocals. They can be used to more carefully examine something in an experiment.
Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M
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
It is referred as a covalent bond
