An accepted and universally true explanation of observed facts is a law.
For the concentration of Na in M when it begins to crystallize out of the solution is mathematically given as
Na+ = 0.125M
<h3>what is the concentration of Na in M when it begins to crystallize out of the solution?</h3>
Generally, the equation for the Ksp of the solution is mathematically given as
Na⁺+CL⁻
Ksp = SxS
53.9 = s^2
S = 7.34
Molar mass Nacl 58,44
Solubility = 7.34/58.44
Solubility = 0.125M
In conclusion, the concentration of Na in M when it begins to crystallize out of the solution
Na+ = 0.125M
Read more about Ksp
brainly.com/question/7185695
Answer:
1.125 moles of H2S
Explanation:
From 2HCl(aq) + ZnS(s) ---------> H2S (g) + ZnCl2 (aq)
From the reaction equation;
2 moles of HCl produces 1 mole of H2S
Therefore 2.25 moles of HCl will produce 2.25 ×1/2 = 1.125 moles of H2S
Recall that it was explicitly stated in the question that ZnS is the reactant in excess. This implies that HCl is the limiting reactant and controls the amount of product obtained.
Answer:
<h2>106.62 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 83.7 g
density = 0.785 g/mL
We have
We have the final answer as
<h3>106.62 mL</h3>
Hope this helps you
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
