First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl
since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol
1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.
8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂
2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl
3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl
3) lastly, we convert the moles of HCl to grams using the molar mass.
molar mass of HCl= 1.01 + 35.5= 36.51 g/mol
7.1x10⁻⁴ mol HCl (36.51 g/mol)=<span> 0.026 grams HCl</span>
Solution :
For the reaction :
we have
![$Ka = \frac{[\text{Tris}^- \times H_3O]}{\text{Tris}^+}$](https://tex.z-dn.net/?f=%24Ka%20%3D%20%5Cfrac%7B%5B%5Ctext%7BTris%7D%5E-%20%5Ctimes%20H_3O%5D%7D%7B%5Ctext%7BTris%7D%5E%2B%7D%24)
Clearing 
, we have 
So to reach 
, one must have the 
concentration of the :
![$\text{[OH}^-]=10^{-pOH} = 6.31 \times 10^{-7} \text{ moles of base}$](https://tex.z-dn.net/?f=%24%5Ctext%7B%5BOH%7D%5E-%5D%3D10%5E%7B-pOH%7D%20%3D%206.31%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctext%7B%20moles%20of%20base%7D%24)
So we can add enough of 1 M NaOH in order to neutralize the acid that is calculated above and also adding the calculated base.
Volume NaOH 
Tris mass 
Now to prepare the said solution we must mix:
gauge to 1000 mL with water.
The atoms didn't change we only added 2 different ones together. 2nd one is right
Answer: Cycling of carbon in the
ocean is also expected As water moves through and across soils, it carries
valuable nutrients.
