Chemistry, 22.06.2019 12:30 kaliyab191

Sodium sulfate dissolves as follows: na2so4(s) → 2na+(aq) + so42- (aq). how many moles of na2so4 are required to make 1.0 l of solution in which the na concentration is 0.10 m?

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Answer from: kathrynway2715

The number of moles of Na_2SO_4 is 0.05 moles.

Explanation:

To calculate the molarity of solution, we use the equation:

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

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

$0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol$

The chemical reaction for the ionization of sodium sulfate follows the equation:

$Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)$

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = $\frac{1}{2}\times 0.1=0.05moles$ of sodium sulfate.

Hence, the number of moles of Na_2SO_4 is 0.05 moles.

Answer from: cmanglona

Number of moles of Na_2SO_4 are required to make 1.0 L of solution is 0.05 mole.

Explanation:

Given:

$Na_2SO_4(s) \ gives\ 2Na+(aq) + SO_4^2- (aq)$

Na concentration=0.10 M.

To find

The number of moles of Na_2SO_4 which required to make 1.0 L of solution.

To find the mole value,using the molarity formula:

Molarity

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

As we have to find the moles of the solute but we know the molarity so,

Molarity of sodium is 0.10 M.

Therefore,

$0.10 \mathrm{~mol} =\frac{\text { Moles of sodium }}{1.0 }$

Simplified we get,

Moles of sodium is 0.10 M.

As we know that, 2 moles of sodium ions are formed by 1 mole of sodium sulfate

Divide the moles of sodium by 2,

=0.10/2

$=0.5\ mole$

Learn more about molarity, refer:

Answer from: Quest

answer: a chemical reaction

explanation:

im not wrong

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

Electrons are transferred from atoms of to atoms of this transfer makes the sodium atoms and the

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Sodium sulfate dissolves as follows: na2so4(s) → 2na+(aq) + so42- (aq). how many moles of na2so4 ar...