There is 1 OH- in 1 molecule of NaOH. 
Also, there is 1 H+ in 1 molecule of lactic acid.
So the reaction is simple.
so just equate the moles
moles of OH- in NaOH = moles of H+ in lactic acid
26.8 x 0.07 = 250 x Mole of lactic
Moles of lactic = 0.0075
so mass = 0.0075 x 90.8 =  0.681 g
        
             
        
        
        
It is important to use the same balance throughout the entire experiment since the calibration of each balance is not the same and changing balances could result in a systematic error. 
There are three types of errors that could affect the results of the experiment. The effect of random or indeterminate errors is hard to predict, its effect on the results of the experiment could be different every time. The second type of error is the systematic or determinate error, which causes a shift in results in a specific direction. The last type of error in an experiment is human error. 
The type of error that could be related to the use of different balances throughout the experiment is the systematic error. Instruments could be a source of error especially if they are poorly calibrated. Also, analytical balances are calibrated differently which may result in inaccuracy in the weighing of chemicals.
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The average atomic mass of the element is the sum of the products of the percentage abundance of isotope and its mass number. Therefore, for atomic mass equal to 58.933, the most abundant isotope is cobalt-59. Thus, the answer is letter C. 
        
             
        
        
        
First, we need to get moles of NaOH:
when moles NaOH = volume * molarity 
                                  = 0.02573L * 0.11 M
                                 = 0.0028 moles 
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
 ∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
               = 9.4 x 10^-4 / 0.034 L
               = 0.028 M
        
                    
             
        
        
        
You can stop the burning of methane with water or carbon dioxide extinguishers but problems arise when you try to use this to stop the burning of the magnesium.
Explanation:
To burn magnesium (Mg) and methane (CH₄) you need to react them with oxygen:
2 Mg (s) + O₂ (g) → 2 MgO + heat
CH₄ (g) + 2  O₂ (g) → CO₂ (g) + 2 H₂O (g) + heat
However at that temperatures magnesium (Mg) is able to react with water (H₂O) and carbon dioxide (CO₂).
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
2 Mg (s) + CO₂ (g) → 2 MgO (s) + C (s)
So the safe option to stop the burning of the magnesium is to limit the oxygen in the air.
we have used the following notations:
(s) - solid
(g) - gas
(l) - liquid
Learn more about:
combustion reactions
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