A chemistry graduate student is given of a diethylamine solution. Diethylamine is a weak base with . What mass of should the student dissolve in the solution to turn it into a buffer with pH ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to significant digits.

65 g (C₂H₅)₂NH₂Br

Explanation:

It seems your question lacks the values required to solve the problem. However, an internet search tells me these are the values (if your values are different, keep those in mind when solving the problem, but the methodology remains the same):

" A chemistry graduate student is given 125 mL of a 0.90 M diethylamine ((C₂H₅)₂NH) solution. Diethylamine is a weak base with Kb=1.3×10⁻³. What mass of (C₂H₅)₂NH₂Br should the student dissolve in the (C₂H₅)₂NH solution to turn it into a buffer with pH=10.53? You may assume that the volume of the solution doesn't change the (C2H5)2NH2Br is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "

To solve this problem we use the Henderson-Hasselbach equation:

We calculate pKa from Kb:

Now we possess all the required data to calculate the concentration of the bromide salt (C₂H₅)₂NH₂Br :

Now we use the total volume to calculate the moles of (C₂H₅)₂NH₂Br required to have that molar concentration:

Finally we calculate the mass of (C₂H₅)₂NH₂Br:

Which rounding to 2 significant figures is 65 g (C₂H₅)₂NH₂Br

b is the correct answer

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explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.