You have to start listing from the bottom :
3. Secondary Consumers
2. Primary Consumers
1. Producer
Answer:
The mass of CH4 is 60, 29 grams.
Explanation:
We use the weight of the atoms C and H for calculate the molar mass:
Weight of CH4= weight C+ 4 x weight H= 12,01 g/mol +4 x 1,008g/mol=
Weight of CH4 =16, 042 g/mol
1molCH4-----16, 042grams
3,758 mol CH4--X= (3,758 mol CH4 x 16, 042 grams)/1 mol CH4=60,285836 grams
Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have

= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is

Number of moles of 
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of 
= 0.167 M
Now the ICE table :

I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,



= 
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)


So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= 
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
Answer:
1. 0.073L
2. 0.028L
3. 0.014L
Explanation:
The volume for the different solutions are obtained as shown below:
1. Mole = 0.53mol
Molarity = 7.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.53/7.25
Volume = 0.073L
2. 0.035mol from a 1.25M
Mole = 0. 035mol
Molarity = 1.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.035/1.25
Volume = 0.028L
3. Mole = 0.0013mol
Molarity = 0.090M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.0013/0.090
Volume = 0.014L