Answer : The volume of 6M NaOH stock solution is, 16.7 mL
Explanation :
To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
where,
are the molarity and volume of NaOH stock solution.
are the molarity and volume of NaOH.
We are given:
![M_1=6M\\V_1=?\\M_2=0.2M\\V_2=500mL](https://tex.z-dn.net/?f=M_1%3D6M%5C%5CV_1%3D%3F%5C%5CM_2%3D0.2M%5C%5CV_2%3D500mL)
Putting values in above equation, we get:
![6M\times V_1=0.2M\times 500mL\\\\V_1=16.7mL](https://tex.z-dn.net/?f=6M%5Ctimes%20V_1%3D0.2M%5Ctimes%20500mL%5C%5C%5C%5CV_1%3D16.7mL)
Thus, the volume of 6M NaOH stock solution is, 16.7 mL
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Woof. Hope it helps!!! :) ....Reason*
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
![Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BMass%20of%20solvent%7D%7D)
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
![0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}](https://tex.z-dn.net/?f=0.565mol%2Fkg%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NaBr%7D%5Ctimes%201000%7D%7B103g%2Fmole%5Ctimes%20315g%7D)
![\text{Mass of NaBr}=18.3g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20NaBr%7D%3D18.3g)
Thus, the mass of sodium bromide added should be, 18.3 grams.