The concentration of the sodium chloride would be 0.082 M
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.
Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles
Equivalent mole of NaCl = 0.325 moles.
Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Molality can be expressed by moles of solute over
kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl
is equal to 0.25 moles of NaCl over 1 kg of water.
Answer:
D. Propanol
Explanation:
C3H7OH the presence of alcohol functional group makes it propanol
Repeatable
If it’s repeatable and you get the same answer constantly then it’s a good and accurate experiment
