Answer:
36.92 mg of oxygen required for bio-degradation.
Explanation:

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )
Moles benzene =
According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.
Then 0.0003846 mol of benzene will react with:
 of oxygen gas
 of oxygen gas
Mass of 0.0011538 moles of oxygen gas:
0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg
36.92 mg of oxygen required for bio-degradation.
 
        
             
        
        
        
Answer:
The answer is " "
"
Explanation:
Given:
Molarity= number of moles
because it is 1 Liter

therefore, 
it takes 20 mL of Tris.

                      
So, take 
 
        
             
        
        
        
Yup go this website for more information http://dwb.unl.edu/calculators/activities/BalEqn.html
        
             
        
        
        
<h3>1.<u> Answer;</u></h3>
False 
<h3><u>Explanation;</u></h3>
Bases have some of the following properties;
- They have a bitter taste 
- They have a slimy, or soapy feel on fingers 
- Most bases react with acids and precipitate salts.
- Strong bases may react violently with acids.
- Bases turn red litmus paper blue
<h3>2. <u>Answer;</u></h3>
An acid 
<h3><u>Explanation;</u></h3>
- When acids are dissolved in water, the concentration of the acid decreases and it becomes dilute. 
- It dissociates in water to give H+ ions or hydrogen ions.
- All acidic solutions contain more hydrogen ions than hydroxide ions, therefore when added to water it increases the concentration of H+ ions in water, as water is a neutral substance whose concentration of H+ ions is equal to OH-.
 
        
        
        
Moles   =  n  = 3.91 mol
                  Pressure  =  P  =  5.35 atm
                  Temperature  =  T  =  323 K
                  Volume  =  V  =  ?
Formula used: Ideal Gas Equation is used,
                                        P V  = n R  T
Solving for V,
                                           V  =  n R T / P
Putting Values,
                           V  =  (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm
                         V  =  19.36 L