Here’s all of them: kingdom, phylum, classes, order, families, genus, and species
        
                    
             
        
        
        
Answer:
The answer to your question is below
Explanation:
a)    HCl  0.01 M
       pH = -log [0.01]
       pH = - (-2)
       pH = 2
b)    HCl = 0.001 M
       pH = -log[0.001]
       pH = -(-3)
       pH = 3
c)    HCl = 0.00001 M
        pH = -log[0.00001]
        pH = - (-5)
        pH = 5
d) Distilled water
        pH = 7.0
e) NaOH = 0.00001 M
        pOH = -log [0.00001]
        pOH = -(-5)
        pH = 14 - 5
        pH = 9
f)  NaOH = 0.001 M
       pOH =- log [0.001]
       pOH = 3
       pH = 14 - 3
       pH = 11
g)   NaOH = 0.1 M
        pOH = -log[0.1]
        pOH = 1
        pH = 14 - 1
        pH = 13
 
 
        
             
        
        
        
Answer:
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Explanation:
Given:
Coulomb meter = -1 CM
Find:
In debye
Computation:
We know that,
1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye
So,
-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye
-1 Coulomb meter = -2.997 × 10²⁹ Debye
 
        
             
        
        
        
Alloys are supposed to give greater strength to metals, which is why gold is mixed with others to make it harder. They have greater strength and are more resistant to erosion.
        
             
        
        
        
Answer:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles] 
0.03 = log [A-] / [0.0500moles] 
1.017152 = [A-] / [0.0500moles] 
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) = 
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>