If the crucible wasn't covered with a lid the reactants may have produced a gas that was released into the surroundings, or mass may have been lost in the form of water vapour.
<span>Report your numerical answer in units of nm. Use significant figur</span>
Scientists make hypothesis in order to make an educated guess on the outcome of the experiment.
The mass of reacted magnesium chloride is 23.75g, percent by mass of solution magnesium chloride is 90.9%.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of MgCl₂ = 200g / 95g/mol = 2.1mol
Moles of NaOH = 20g / 0.5mol
Given chemical reaction is:
MgCl₂ + 2NaOH → 2NaCl + Mg(OH)₂
From the stoichiometry of the reaction it is clear that
- 1 mole of MgCl₂ = reacts with 2 moles of NaOH
- 0.5 mole of NaOH = reacts with (1/2)(0.5)=0.25 moles of MgCl₂
Mass of reacted MgCl₂ = (0.25mol)(95g/mol) = 23.75g
Percent by mass of MgCl₂ in the given solution mixture will be calculated as:
- % by mass = (Mass of MgCl₂ / Total mass of mixture) × 100%
- Percent by mass of MgCl₂ = (200/200+20)×100% = 90.0&
In the above reaction we obtain NaCl as a solid, and MgCl₂ is the limiting reagent in it, from which 2 moles of NaCl is produced means
- 0.25 moles of MgCl₂ = produces 0.5 moles of NaCl
Mass of NaCl = (0.5mol)(58.5g/mol) = 29.25g
Hence, mass of reacted MgCl₂ is 23.75g, percent by mass of solution magnesium chloride is 90.9% and mass of the obtained solid is 29.25g.
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Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
