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.
Answer:
Oxygen
Explanation:
because oxygen is used when making and sustaining a fire.
I think this seems like something is missing from the question
Answer:
Explanation:
Let the number of moles of oxygen = x
2H2 + O2 --> 2 H2O
x 13.3
Since the balance number for oxygen is 1 and the balance number for water is 2, you must set up a proportion. (Those balance numbers represent the number of moles).
1/x = 2 / 13.3 Cross Multiply
2*x = 13.3 Divide both sides by 2
2x/2 = 13.3/2
x = 6.65
You need 6.65 moles of oxygen.
Answer:
26 grams of D will be produced.
Explanation:
The reaction is given by:
A + B -----> C + D
Mass of A reacted = 21 g
Mass of B reacted = 22 g
Mass of C formed = 17 g
Mass of D formed = m =?
According to law of conservation of mass, the total mass of the reactants used is equal to the total mass of the product formed.
Then:
mass of A reacted + mass of B reacted = mass of C formed + mass of D formed
21 + 22 = 17 + m
m = 26 g
Answer:

Explanation:
The expression for the work done is:

Where,
W is the amount of work done by the gas
R is Gas constant having value = 8.314 J / K mol
T is the temperature
P₁ is the initial pressure
P₂ is the final pressure
Given that:
T = 300 K
P₁ = 10 bar
P₂ = 1 bar
Applying in the equation as:



