The water molecule<span> has two </span>hydrogen<span> atoms joined to one </span>oxygen<span> atom by single covalent bonds. </span>Because oxygen is more electronegative than hydrogen<span>, the electrons of the covalent bond spend </span>more <span>time closer to the </span>oxygen<span> then </span>hydrogen<span> which creates a polar covalent bond. Hope this answers your question. Have a great day!</span>
Answer : The pH of the solution is, 4.9
Explanation : Given,
Dissociation constant for acetic acid = 
Concentration of acetic acid = 0.05 M
Concentration of sodium acetate = 0.075 M
First we have to calculate the value of 
.
The expression used for the calculation of is,
Now put the value of 
in this expression, we get:
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of the solution is 4.9.
<span> indefinite volume </span>
<span>If the mass is definite and the density is definite, </span>
<span>then your volume is definite.</span>
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
