Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for 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.
Answer:
1.3×10⁻³ M
Explanation:
Hello,
In this case, given the dissociation reaction of acetic acid:
We can write the law of mass action for it:
![Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BCH_3CO_2%5E-%5D%7D%7B%5BCH_3CO_2H%5D%7D)
Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change 
due to the dissociation extent, we are able to rewrite it as shown below:
Thus, via the quadratic equation or solve, we obtain the following solutions:
Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.
Regards.
