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.
Explanation:
From the knowledge of law of multiple proportions,
mass ratio of S to O in SO:
mass of S : mass of O
= 32 : 16
= 32/16
= 2/1
mass ratio of S to O in SO2:
= mass of S : 2 × mass of O
= 32 : 2 × 16
= 32/32
= 1/1
ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:
= 2/1 ÷ 1/1
= 2
Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.
We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
Answer:
the new pressure is 2.09 atm
Explanation:
you have to use gay lussac's law so the formula is
p1/t1 = p2/t2
and convert C to Kelvin k=C+273.15
1.72atm/294.15 = p2/358.15
solve for p2 by multiplying 358.15 on both sides
p2=2.09 atm
