Answer:
Aqueous sodium ions, Aqueous chloride ions
Explanation:
Sodium chloride dissociates into Na+ and Cl- ions in water
Sodium chloride crystals may also be present in case of saturated solution
It is practical research of Boyle’s law..
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Answer:
V₂ = 12.5 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial pressure = 50.0 KPa
Final volume = ?
Final pressure = 400.0 KPa
Solution:
The given problem will be solved through the Boyle's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
50.0 KPa × 100.0 mL = 400 KPa × V₂
V₂ = 5000 KPa.mL/ 400 KPa
V₂ = 12.5 mL
Answer :
(a) The number of moles of D produced can be, 6.67 moles.
(b) The volume of D prepared can be, 24.5 L
Explanation :
The given chemical reaction is:
Part (a) :
From the balanced chemical reaction, we conclude that:
As, 3 moles of A react to give 5 moles of D
So, 4 moles of A react to give 
moles of D
Thus, the number of moles of D produced can be, 6.67 moles.
Part (b) :
As we know that 1 moles of substance occupies 22.4 L volume of gas.
As, 
volume of B gives 
volume of D
As, 9.8 L volume of B gives 
volume of D
Thus, the volume of D prepared can be, 24.5 L
Answer:
1019.27 g.
Explanation:
- For the balanced reaction:
<em>2Na + Cl₂ → 2NaCl,</em>
It is clear 2 moles of Na with 1 mole of Cl₂ to produce 2 moles NaCl.
- Firstly, we need to calculate the no. of moles of Cl₂ is needed to react with 57.5 mol Na:
2 moles of Na need → 1 mol of Cl₂, from the stichiometry.
57.5 moles of Na need → ??? mol of Cl₂.
<em>∴ The no. of moles of Cl₂ is needed to react with 57.5 mol Na =</em> (1 mol)(57.5 mol)/((2 mol) <em>= 28.75 mol.</em>
<em>∴ the mass of Cl₂ is needed to react with 57.5 mol Na = (no. of moles of Cl₂)(molar mass of Cl₂) =</em> (28.75 mol)(35.453 g/mol) <em>= 1019.27 g.</em>
