Chemistry, 26.03.2020 20:02 Cassandoo

A 0.629 g sample of a diprotic acid is dissolved in water and titrated with 0.270 M NaOH. What is the molar mass of the acid if 36.4 mL of the NaOH solution is required to neutralize the sample? Assume the volume of NaOH corresponds to the second equivalence point. A flask with a solution sits on the base of a ring stand. A buret filled with liquid is suspended above the flask by the ring stand. molar mass: g/mol

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Answer from: manillasoccer

The molar mass of acid is 127.97 g/mol

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of NaOH solution = 0.270 M

Volume of solution = 36.4 mL

Putting values in above equation, we get:

$0.270M=\frac{\text{Moles of NaOH}\times 1000}{36.4}\\\\\text{Moles of NaOH}=\frac{0.270\times 36.4}{1000}=0.00983mol$

The chemical equation for the reaction of NaOH and diprotic acid follows:

$2NaOH+H_2X\rightarrow 2NaX+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

So, moles of diprotic acid = $\frac{0.00983}{2}=0.004915moles$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of acid = 0.629 g

Moles of acid = 0.004915 moles

Putting values in above equation, we get:

$0.004915mol=\frac{0.629g}{\text{Molar mass of acid}}\\\\\text{Molar mass of acid}=\frac{0.629g}{0.004915mol}=127.97g/mol$

Hence, the molar mass of acid is 127.97 g/mol

Answer from: Quest

step-by-step explanation:

whole numbers, zero or greater

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.