Answer:
All of the above are true
Explanation:
a) The emission spectrum of a particular element is always the same and can be used to identify the element: It's true since the emission spectrum for each element is unique. It has the same bright lines at the same wavelength. This feature is used to identify elements. For example, the study of the emission spectra of light arriving from stars allow us to identify the elements presents in the star because the light contains the emission spectra of those elements.
b)The uncertainty principle states that we can never know both the exact location and speed of an electron: It is true since the velocity of an electron is related to its wave nature, while its position is related to its particle nature and we cannot simultaneously measure electron's position and velocity with precision.
c) An orbital is the volume in which we are most likely to find an electron: An orbital is a probability distribution map that is used to decribe the likely position of an electron in an atom.
B) False- It has seven
A hexagon would have 6.
Explanation:
To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23.
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.
Answer is: concentration of hydrogen iodide is 6 M.
Balanced chemical reaction: H₂(g) + I₂(g) ⇄ 2HI(g).
[H₂] = 0.04 M; equilibrium concentration of hydrogen.
[I₂] = 0.009 M; equilibrium concentration of iodine.
Keq = 1·10⁵.
Keq = [HI]² / [H₂]·[I₂].
[HI]² = [H₂]·[I₂]·Keq.
[HI]² = 0.04 M · 0.009 M · 1·10⁵.
[HI]² = 36 M².
[HI] = √36 M².
[HI] = 6 M.
