Bc an MRI is just a scanner it's not and never will be meant for treatment
Answer:0.026ml
Explanation:
Details are found in the image attached. We must subtract the saturated vapour pressure of hydrogen gas at the given temperature from the total pressure of the hydrogen gas collected over water to obtain the actual pressure of hydrogen gas and substitute the value obtained into the general gas equation. The dry hydrogen gas has no saturated vapour pressure hence the value is substituted as given. All temperatures must be converted to Kelvin before substitution.
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
The question is incomplete. The complete question is :
Hydrogen is manufactured on an industrial scale by this sequence of reactions:
The net reaction is :
Write an equation that gives the overall equilibrium constant 
in terms of the equilibrium constants 
and 
. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
Solution :
![$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%3D%20%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%24)
...............(1)
![$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_2%20%3D%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
...................(2)
![$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
On multiplication of equation (1) and (2), we get
![$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$](https://tex.z-dn.net/?f=%24K_1%20%5Ctimes%20K_2%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5Ctimes%20%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%24)
![$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$](https://tex.z-dn.net/?f=%24K_1K_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%5E4%7D%7B%5BCH_4%5D%5BH_2O%5D%5E2%7D%24)
.................(4)
Comparing equation (3) and equation (4), we get
