Answer:
1.
A= <u>sum</u><u>(</u><u>mass</u><u>*</u><u>percent</u><u> </u><u>abundance</u><u>)</u>
M 100
=(23.985*78.70)+(24.946*10.13)+(25.983*11.17)/100
= 24.3
2. The element is Magnesium.
3. 2412Mg,2512Mg and 2612Mg
To find - Identify what kind of ligand (weak or strong), what kind
of wavelength (long or short), what kind of spin (high spin or
low spin) and whether it is paramagnetic or diamagnetic for
the following complexes.
1. [Mn(CN)6]4-
2. [Fe(OH)(H2O)5]2
3. [CrCl4Br2]3-
Step - by - Step Explanation -
1.
[Mn(CN)⁶]⁴⁻ :
Ligand - Strong
Wavelength - Short
Spin - Low spin
Number of unpaired electrons = 1 ∴ paramagnetic.
2.
[Fe(OH)(H₂O)₅]²⁺ :
Ligand - Weak ( both OH⁻ and H₂O )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 5 ∴ paramagnetic.
3.
[CrCl₄Br₂]³⁻ :
Ligand - Weak ( both Br⁻ and Cl⁻ )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 3 ∴ paramagnetic.
Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] =
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] =
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] =
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] =
[In-]/[HIn] = 63.10
Answer:
the answer to this is O hypothesis
Answer:
picograms
Explanation:
Please see attached picture for full solution.