Answer:
a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.
b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.
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:
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
Explanation:
Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:
93.7 g of C
6.3 g of H
Let's make a rule of three:
In 100 g of naphthalene we have 93.7 g of C and 6.3 g of H
In 128 g of naphthalene we would have:
128 . 93.7 / 100 = 120 g of C
128. 6.3 / 100 = 8 g of H
We convert the mass to moles, by molar mass:
120 g . 1mol / 12 g = 10 moles C
8 g . 1mol/ 1g = 8 moles H
Molecular formula naphthalene → C₁₀H₈
Empirical formula naphthalene → C₅H₄
(The sub-index of each element is divided by the largest possible number)
Answer:
Explanation:
A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features. The t-test is one of many tests used for the purpose of hypothesis testing in statistics. Calculating a t-test requires three key data values.
Answer :
The correct answer for primary component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .
<u>Buffer solution :</u>
It is a solution of mixture of weak acid and its conjugate base OR weak base and its conjugate acid . It resist any change in solution when small amount of strong acid or base is added .
<u>Capacity of a good buffer : </u>
A good buffer is identified when pH = pKa .
From Hasselbalch - Henderson equation which is as follows :
![pH = pka + log \frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pka%20%2B%20log%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
If [A⁻] = [HA] ,
pH = pka + log 1
pH = pKa
This determines that if concentration of weak acid and its conjugate base are changed in small quantity , the capacity of buffer to maintain a constant pH is greatest at pka . If the amount of [A⁻] or [HA] is changed in large amount , the log value deviates more than +/- 1M and hence pH .
Hence Buffer has best capacity at pH = pka .
<u>Phosphate Buffer : </u>
Phosphate may have three types of acid-base pairs at different pka ( shown in image ).
Since the question is asking the pH = 7.4
At pH = 7.4 , the best phosphate buffer will have pka near to 7.4 .
If image is checked the acid - base pair " H₂PO₄⁻ and HPO₄²⁻ has pka 7.2 which is near to pH = 7.4 .
Hence we can say , the primary chemical component of phosphate buffer at pH = 7.4 is H₂PO₄⁻ and HPO₄²⁻ .
