1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Science of, or study of contours, or the lines outlining a scene, drawing, etc.
Answer: Dissociation constant of the acid is
.
Explanation: Assuming the acid to be monoprotic, the reaction follows:

pH of the solution = 6
and we know that
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=antilog(-pH)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-pH%29)
![[H^+]=antilog(-6)=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dantilog%28-6%29%3D10%5E%7B-6%7DM)
As HA ionizes into its ions in 1 : 1 ratio, hence
![[H^+]=[A^-]=10^{-6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BA%5E-%5D%3D10%5E%7B-6%7DM)
As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:
![[HA]=[HA]-[H^+]](https://tex.z-dn.net/?f=%5BHA%5D%3D%5BHA%5D-%5BH%5E%2B%5D)
![[HA]=0.1M-10^{-6}M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.1M-10%5E%7B-6%7DM)
![[HA]=0.09999M](https://tex.z-dn.net/?f=%5BHA%5D%3D0.09999M)
Dissociation Constant of acid,
is given as:
![K_a=\frac{[A^-][H^+]}{HA}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7BHA%7D)
Putting values of
in the above equation, we get


Rounding it of to one significant figure, we get

The pH of a 0.0115 m aqueous formic acid solution is mathematically given as
pH=2.8424
This is further explained below.
<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>
Generally, the equation for the chemical equation is mathematically given as
HCOOH H^+ + HCOO


![&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424](https://tex.z-dn.net/?f=%26%5Cleft%5B%5Cmathrm%7BH%7D%5E%7B%2B%7D%5Cright%5D%3D%5Cmathrm%7BC%20%5Calpha%7D%5C%5C%5C%5C%26%3D0.125%20%5Ctimes%200.0115%5C%5C%5C%5C%26%3D1.4375%20%5Ctimes%2010%5E%7B-9%7D%5C%5C%5C%5C%26P%3D-%5Clog%20%5Cleft%5BH%5E%7B%2B%7D%5Cright%5D%5C%5C%5C%5C%26%3D-%5Clog%20%5Cleft%5B1.4375%20%5Ctimes%2010%5E%7B-3%7D%5Cright%5D%5C%5C%5C%5C%26P%20H%3D2.8424)
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B cause it tells how it moves