Answer:
![M=1.62M](https://tex.z-dn.net/?f=M%3D1.62M)
Explanation:
Hello.
In this case, since the molarity is computed by the division of the moles of the solute by the volume of the solution in liters:
![M=\frac{n_{solute}}{V_{solution}}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7Bn_%7Bsolute%7D%7D%7BV_%7Bsolution%7D%7D)
We first compute the moles of glucose (molar mass 180 g/mol) as shown below:
![n_{solute}=582.0g*\frac{1mol}{180g} =3.233mol](https://tex.z-dn.net/?f=n_%7Bsolute%7D%3D582.0g%2A%5Cfrac%7B1mol%7D%7B180g%7D%20%3D3.233mol)
Therefore, the molarity turns out:
![M=\frac{3.233mol}{2.00L}\\ \\M=1.62M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B3.233mol%7D%7B2.00L%7D%5C%5C%20%5C%5CM%3D1.62M)
Best regards.
Hydrocarbons are derived from the elements, Hydrogen and Carbon. Therefore hydrocarbons contain both carbon and hydrogen.
Sodium chloride because it contains the most reactive metal(sodium) and most reactive non-metal(chlorine).
10 mL of 0.1 M NaOH is equal to 1 equivalent of base, if titrating 10.00 mL of 0.1 M amino acid.
Given :
0.1 M NaOH
0.1 M Amino acid
10.00 mL volume of Amino acid
<em> R-CH(NH₂)-COOH + NaOH → R-CHNH₂Na + H₂O</em>
In this case, end factor = 1 and It is 1 mole / 1 mole ratio
Since, the concentration of amino acid and NaOH are same, that is 0.1 M.
Concentration of amino acid = concentration of NaOH
0.1 M = 0.1 M
Therefore, 10 mL of NaOH is required to neutralize 10 mL of amino acid and the end factor is equivalent to 1.
Hence, The volume of 0.1 M NaOH is equal to 1 equivalent of base.
To learn more about titration,
brainly.com/question/14151679
#SPJ4